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About Google Book Search Google's mission is to organize the world's information and to make it universally accessible and useful. Google Book Search helps readers discover the world's books while helping authors and publishers reach new audiences. You can search through the full text of this book on the web at |http: //books .google .com/I / J t ' ^y^ -j ^ i- ^, f~ L .-y.r.j^.^. ^' 1 ^^^fl^'^^Aidli ■^ • * - /^f^'^/i ii -M: . Cq Ay.^x /Mn /a^ %^i /■y-^ // ^ i P'' 4'. <-//-_' 1^^,^^ -6 /- A '^7. c ^^6j t>^ ,lJ^ y ^i ^ i- -^ y .^ \^lii^/a&L5x^ ^ V' //I ^?^ £>^tj r ADAMS'S NEW ARITHHETIG— RBTISED BDITIOH. ARITHMETIC, III WHICH THI PRINCIPLES OF OPERATING BY NUMBERS ANALYTICALLY EXPLAINED Am SYNTHETICALLY APPLIED. ILLUSTRATED BY COPIOUS EXAMPLES DSSIGNSD FOR THE USB OF SCHOOLS AND ACADEMIES. BY DANIEL ADAMS, M. D., AUTHOS OP THI SCHOLAK's ABITBMBTIC, SCHOOL aBOGRAPHT, BTO BOSTON: PUBLISHED BY PHILLIPS & SAMPSON. 1S4S. A3 oil GIFT OF Erfitered according to Act of Cohgress, in the year 1848, by DANIEL ADAMS, M. D. In the Clerk's Office of the District Court of the District of New Hampshire. EDUCATION OEPT Steraotjped by OEOROE A. CURTIS; MSW BNOULND T 'PB AND STBRBOTTPB VOTOIMIT BOSTON '( PBEFACE. TBI *< Scho ar's Arithmeuc," by the author of the present work, was first published in 1801. The great favor with which it was received is an evidence tUat it was adapted to the wants of schools at the time. At a subseflttent period the analytic method of instruction was applied to arithmetic, with much ingenuity and success, by our late lamented countr3rman, Wakren Colburn. This* was the great improvement in the modem method of teaching arithmetic. The author then 3rielded to the solicitations of numerous fiiends of education, and prepared a work combining the analytic with the synthetic method^ which was published m 1827, with the title of "Adams' New Arithmetic." Few works ever issued from the American press have acquired so great popularity as the " New Arithmetic." It is almost the only work on arithmetic used in extensive sections of New England. It has been re-published in Canada, and adapted to the currency of that province. It has been translated into the language of Greece, and published in that country. It has found its way into every part of the United States. In the state of New York, for example, it is the text-book in ninety- three of the one hundred and fifly.five academies, which reported to the regents of the University in 1847. And, let it be remarked, it has se- cured this extensive circulation solely by its merits. Teachers, super- intendents, and committees have adopted it because they have found it fitted to its purpose, not because hired agents have maae unfair repre- sentations of its merits, and, of the defects of other works, seconding their arguments by liberal pecuniary offers — a course of dealing re- cently introduced, as unfair as it is injurious to the cause of education. The merits of the "New Arithmetic" have sustained it very success- fully against such exertions. Instances are indeed known, in which it has been thrown out of schools on account of the " liberal ofiers" of those interested in other works, but has subsequently been readopted without any efforts from its publishers or author. The "New Arithmetic" was the pioneer in the field which it has occupied. It is not strange, then, that teachers should find defects and deficiencies in it which they would desire to see removed, though they might not think that they Would be profited by exchanging it for any other work. The repeated calls of such have induced the author to un- dertake a revision, in which labor he would present acknowledgments to numerous friends for important and valuable suggestions. Mr. J. HoMsa French, of Phelf^, N. Y., well known as a teacher, has been engaged with the author in this revision, and has rendered important aid. Mr. W. B. Bunnell, also, principal of Yates Academy, N. Y., formerly principal of an academy m Vermont, has assisted throughout the work, having prepared many of the articles. The revision after Fercentage is mostly his work. TV PRXFACB. The characteristics of the << New Arithmetic,'' which hare given the work so great popularity, are too well known to require any notice here. These, it is believed, will be found in the new work in an improved form. One of the peculiar characteristics of the new -vp^rk is a more natural and philosophical arrangement. After the consideration of simple whole numoers, that of simple fractional numbers should evidently be intro- duced, since a part of a thin^^ needs to be considered quite as frequently as a whole thing. Again ; since the money unit of the federal currency is divided decimally. Federal Money certainly ought not to precede Decimal Fractions. It has been thought best to consider it in connec- tion with decimals. Then follow Compound Numbers, both integral and fractional, the reductions preceding the other operations, as thev necessarily must. Percentage is made a general subject, under whidbi are embraced many particulars. The articles on Proportion, Alligation^ and the Progressions will be found well calculated to make pupils thor oughly acquainted with these interesting but difficult subjects. Care has been taken to avoid an arbitrary arrangement, whereby the processes will be purely mechanical to the learner. If, for instance, aJl the reductions in common fractions precede the other operations, the pupil will have occasion to divide one traction by another long before he shall have learned the method of doing it, and must proceed by a rule, to himself perfectly unintelligible. The studied aim has been through- out the entire work to enable the ordinary pupil to understand every thing as he advances. The author is yet to be convinced that mental discipline will be promoted, or any desirable end be subserved, by con- ducting the pupil through blind, mechanical processes. Just so far as he can understand, and no farther, is there prospect of benefit. No good results from presenting things, however excellent in themselves, if they are beyond the comprehension of the learner. Those teachers who prefer to examine their classes by questions, will find that little will escape the pupil's attention, who shall correctly an- swer all those in the present work, while teachers who practise the far superior method of recitation by analysis, will find the work admirably adapted to their purpose. The examples, it is hoped, will require very full applications of the principles. Many antiquated things, which it has been fashionable to copy in arithmetics, from time immemorial, have been omitted or improved, while new and practical matter has been introduced. A Key to this revision is in progress. With these remarks, the work is submitted to the candid examination of the public, by The Author. Keene, N, H,^ Febntary^ 1848. SUGGESTIONS TO TEACHERS. The writer complies with the request of the venerable author of <• Ad- ams' Arithmetic," to preface the new work with a few suggestions lo his associates in the work of instruction. Though he has been engaged for sometime past in assisting to make the work better fitted to ac- complish its design, he is perfectly satistied that improvement in school education is rather to be sought in improved use of the books which we now have, than in making belter books. Better arithmeticians would be made by the book as it was before the present revision, using it as it might be used, than will probably be madt in most cases with the new work, even though the former were very defective, the latter perfect. Exertion, then, to bring teachers to a higher standard, will be more effective in improving school education, than any efforts at improving school books can possibly be. It is here where the great improvement m ust be sought. Without the cooperation of competent teachers, the greatest excellences in any book will remain unnoticed, and unimproved. Pu- pils will frequently complain that they have never found one that could explain some particular thing, of which a full explanation is given in the book which they have ever used, and their attention only needed to have been called to the explanation. Then let teachers make themselves, in the first place, thoroughly acquainted with arithmetic. The idea that they can " study and keep ahead of their classes," is an absurd one. They must have surveyed the whole field in order to conduct inquirers over any part, or there will be liability to ruinous misdirection. Young teachers are little aware of their deficiencies in knowledge, and still less aware of the injurious effects which these deficiencies exert upon pupils, who are often dis- gusted with school education, because they are made to see in it so little that is meaning. In the next place, let no previous famiUarity with the subject excuse teachers from carefully preparing each lesson before meeting their classes. Thereby alone will they feel that freshness of interest, which will awaken a kindred interest among their pupils j and if on any occa- sion they are compelled to omit such preparation, »they will discover a declining of interest with their classes. Teachers who are obliged to have their books open, and watch the page while their classes recite, are anfit for their work. Pupils should be taught how to study. That, after all, is the great object of educating. The facilities for merely acquiring knowledge are abundant, if persons know how to improve them. The members of classes will often fail in recitation, not because they have not tried, but have not known how to get their lesson. They neglect trying, because they can do so little to advantage. They may read over a statement in their book a dozen times, they say, but cannot remember it, — because ikqf da not understand it. An hour spent with each pupil individual I* VI SITGGSSTtONS TO TEACHERS. in questioning him on the meaning of each sentence, which ne may be required to read, will be of incalculable advantage. When pupils shall have been taught how to study, le.t them be re- quired to get their lessonSf and recite them. If the present book is not thought by teachers to contain a sufficient description, and a sufficient explanation of everything, let them try to find one that does, for if pupils present themselves before the blackboard at the time of recitation, with the expectation that the teacher is to explain to the class, and help them through with what they cannot go through themselves, they will not feel that they must have studied themselves ; and the paltry oralizing o* the teacher will not be listened to, or if heard, will not be understood, or at best, not retained in memory. Pupils may be made to see things for the moment, while no abiding impression will remain on their minds. They will often proceed, in such a manner, through a book, and perhaps have the mistaken idea that they understand its contents — to perpetuate the evil of superficialism, perhaps, themselves, a& teachers. Pupils will never have a sufficient understanding of a sub- ject till they shall have studied it carefully themselves, and mastered each part by severe personal application. Recitation by analysis will be found more conducive to thorouga scholarship than adherence to any written questions. Let the class, or any member oi the class, be able to commence at the beginning and go through with the entire lesson without any suggestion from the teacher, — a thing that is perfectly practicable and easily attainable. Let pupils be called on, at the pleasure of the teacher, in any part of the class, to ^o on with the recitation, even to proceed with it in the midst of a sub- ject, the topic in no case ever being named by the teacher. They will thereby become accustomed to give their attention to the recitation, and they will be profited from it, besides securing habits of attention, which will be of incalculable value. In fine, let arithmetic be studied properly, knd more valuable mental discipline will be acquired from it^than is often attained from the whole course in mathematics usually assigned by college faculties. It is not the extent, but the value of acquisitions in mathematics, which is desirable. ^ W. B. B. INDEX. SIMPLE NDMBERa N i*«tion and Nmnent^xi, 9 AAlition, 16 Review of Numeration and Addition, . 22 Subtraction, 23 Review of Subtraction, 29 Multiplication, 31 , ilTnetration by Diagram, 34 Contractions in Multiplication, .... 40 Review of Multiplication, 45 Division, 47 — -, illustration by Diagram, ... 60 Contraction! in DiTl8to% ••..... 69 Review of Division, 63 Miscellaneous Exercises, 65 Problems in the Measurement of Rec- tangles and Solids, 69 , illustration by Diagram, . 71 Definitions. 73 General Principles of Division, .... 74 Cancelation, 75 Common Divisor, 78 Greatest Common Divisor, 78 COMMON FRACTIONS. Notation of Common Fractions, .... 80 Proper, Inrjproper, &c., 82 Reduction of Fractions, 83 I^ reduce a fraction to its lowest terms, 85 Addition and Subtraction of Fractions, . 87 Common Denominator, 87 ' — , 1st method, . .88 , 2d method, . . 89 Least Common Denominator, or Least Common Multiple, 90' New Numerators, 91 General Rule, 91 Multiplication of Fractions, 93 by a whole number, two ways, 94 Multiplication of whole numbers by a fraction, 95 Multiplication of one fraction by another, 96 General Rule, 97 Examples in Cancelation, 9S Division of Fractions, 99 by a whole num- ber, two ways, 100 Division of whole numbers by a fraction, 101 Division of one fraction by another, '. 103 General Rule, 103 Reduction of Complex to Simple Frac- tions, 104 Promiscuous Examples, 106 Review of Common Fractions, .... 106 DECIMAL FRACTIONS AND FEDERAL MONEY. Decimal Fractions, 108 Noution of Decimal Fractions, . . .110 Table, Ill To read Decimals 112 To write Decimals, . % 112 Reduction of Decimal Fractions, . . .113 of Common to Decimal Frac- tions, 114 Federal Money, 116 Reduction of Federal Money, • . . . 1 18 Addition and Subtraction of Decimal Fractions, 119 Addition and Subtraction of Federal Mon2y, 120 Multiplication of Decimal Fractions, 121 , illustration by Diagram, 122 — ^— of Federal Money, . . 123 Division of Decimal Fractions, . . . 124 of Federal Money, 126 Review of Decimal Fractions, . . . . 127 Bills 129 COMPOUND NUMBERS. Definition, 131 Reduction of Compound Numbers, . . 132 ————— English Money, .... 132 WBIOHT. 1. Avoirdupolse Weight, 135 ! n. Troy Weight. .... 136 ■ . . J I I in. Apoibacarr'a weight, 137 MBAStnUI OP BXTBNSION. Reduction of, L Linear Measure, . . 133 aoth Measure, . . . . 13J n. Land, or Square Measure, 140 in. Cubic Measure. . . 141 Tin INDEX* lUUBTmi OF OAFACITT. Reduction of, I. Wine Measure, . . # 143 — — — II. Beer Measure, ... 144 -^— — — — III. Dry Measure, . . . 144 Time, 146 Circular Measure, ... 146 Miscellaneous Table 147 Reduction of Fractional Compound Numbers, 147 To reduce a fraction of a higher de- nomination to one of a lower, . . . 148 To reduce a fraction of a lower to a higher denomination 148 To reduce a fraction of a higher to in- tegers of a lower denomination, . . 149 To reduce integers of a lower tojpc- tions of a higiier denomination, . . 149 Reduction of Decimal Compound Num* bers, 161 Review of Reduction of Compound Numbers, 163 Addition of Com|)ound Numbers. . . 156 - Fractional Compound Num* bers, 160 Subtraction of Compound Numbers, . 160 Subtraction of Fneilooal Oanpoand Numbers, 164 Multiplication and DiTlakm of Com- Ssund Numbers, ......... 165 erence in longitude and lime be* tween diflRbrent places. Diagram, . . 171 Reriew of Compound Numbers, ... 172 Analysis, 174 Given, price of unity, the quantity, to dni tne price of quantity, 176 Given, quantity, price of quantity, to find the price oi unity, 175 Given, price of unitv, pnco of quantity, to find the quantity, i 176 Practice. Aliquot ParU, 178 Articles sold by 100, 180 by the ton of 3000 lbs., ... 181 price, aliquot part of a pound, kc 183 Articles, quantity less than unity, . . 184 To reduce shillings, pence, kc., to the decimal of a pound, 186 To reduce the decimal of a pound to shillings, pence and fiurthingiB, ... 187 PERCENTAGR Definitian, 187. Rule, 188 Insurance, ....... 190 Mutual Insurance, 191 Sioclcs. 193 Brokerage, . 194 Profit and Loss, 194 Interest, 195. General Rule, 199 Easy way of casting interest when the rate is 6 per cent., {iOO To compute interest on pounds, shil* lings, pence, &;c., 205 To compute interest when partial pay- ments have been made, 205 Con}pound Interest, 209 Table, 211 Annual Interest, 212 Time, rate, and amount given, to find the principal, 214 Discount, 21 C. Commission, .... 216 Time, rate, interest, to find the princi* pal, .217 Principal, interest, time, to find the rate, 218 Principal, rate, Interest, to find the time, 218 Percentage to find the rate 219 Bankruptcy, 220 General Average, 221 Partnership, 222 on Time, 223 Banking 224 Taxes, method of assessing, 2S6 Duties, 22? Specific, 228 Ad Valorem, 229 Review of Percentage, 230 Equation of Payments, 233 Ratio 235 Inverted and Direct Ratios, 235 Compound Ratio, 2.36 Proportion, 236 Rule of Three, 237 to invert both Ratios, 238 one Ratio, • . . 239 To find the fcurth term of a proportion when three are given, 239 Cancelation Applied, 240 Compound Proportion, 242 , Rule, 844 Review of Proportion, 245 Alligation Medial, 246 Alternate, 247 Exchange, 251 with England, 253 France, 254 Value of Gold Coins, 255 Duodecimals, 256 , scale for talcing Dimen- sions in feet and Decimals of a foot, 259 Involution, 860 Srolutioa, 868 Extraction of the Square Root 862 Practical Exercises, 266 Extraction of the Cube Root, .... 269 Practical Exercises, 273 Review, 274 Arithmetical Progression, 276 Simple Interest by Progression, . . . 277 Annuities by Arithmetical Pn^ra«sion,2!9 Geometrical Progression, 281 Compound .Interest by Progression, . • 283 ComfMund Discount. — ^Tabto, 885 Annuities at Compound Interest, . . . 288 Present worth of^ Annuities at Com- pound Interest 289 Present worth of Annuities. Table, . 290 ■ in Rever- sion, 291 Perpetual Annuities, . • 898 Permutation 893 Miscellaneous Examples, ...... 894 Measurement of Surfaces, 298 Solids, 300 Guaging, 391 Forms of Notes, Ac, •••••••• 308 Bills, . a04 ARITHMETIC. . * » 4 « tf « * • > • J J^ * W O • « NOTATION AND NUMERATION. ^ 1« A single thing, as a dollar, a horse, a man, &c., ia called a unit, or one» One and one more are called two, two and one more are called three, and so on. Words expressing how many (as one, two, three, &c.) are called numbers. This way of expressing numbers by words would be very slow and tedious in doing business. Hence two shorter meth- ods have been devised. Of these, one is called the Roman* method, by letters; thus, I represents oqe ; V, five; X, ten, &c., as shown in the note at the bottom of the page. The other is called the Arahic method, by certain charao ters, called ^^rcs. This is that in general use. * In the Roman method, by letters, I represents one; W^JLve; X, ten; L, Jifty; C, one hundred; D,jl»c hundred; and M, one thousand. As oAen as any letter is repeated, so many times its value is repeated, un- less it be a letter representing a less number placed before one representing a greater; then the less number is taken from tne greater ; thus, IV represents /bur; IX, rune, &c., as will be seen in the following One Two Three Four Five Six Seven Eight • Nlue- Ten Twenty Thirty Forty Fifty Sixty Seventy Eighty I. II. III. nil. or IV. V. VI. VII. VIII. Vnil. or DC. X. XX. XXX. XXXX. or XL. L. LX. LXX. LXXX. TABL.E. Ninety One hundred Two hundred Three hundred Four hundred Five hundred Six hundred Seven hundred Eight hundred Nine hundred One thousand Five thousand Ten thousand Fifty thousand Hundred thousand One million Two million LXXXX. or XC C. CO. CCC. CCCC. D. or 10.* DC. Dec. DCCC, DCCCC. M. or ClO.t 100. or v.? CCIOO. or X. 1000 CCClboO. orC M. MM". * ID is used instead of D to represent five hundred, and for every additional an nexed at the right hand, the number is increased ten times. t CiO is used to represent one thousand, and for every C and put at each end, th* niunber is increased ten times. t A lino over any number increaaea its valua one thousand tUnns, n 10 NOTATION AND NUMERATION. II 2, 3. In the Arabic method the first nine numbers have each a separate character to represent it ; thus, ^\:^\ L^il'fJ'tt. I Nou I. These nine ch^ Jiii|g,..is repwsai^ ^y this S ^^^ ^ ^,^ sigmjkmt zhQX&cteXf • • • • • I • !• figures, because they each /;J|vtx-^iixteo'V;filsdiaracte?, 2. represent some number. Thi'ee' tir^f^, D}rtiHS^db.^aeter, 3. Sometimes, also, they are Four units, by this character, 4. called digits. Five units, by this character, 5. ^ ^^^f 2. ^ The value of Six units, by this character, 6. ^^^ifi^^f '. ^ ^^, "^''Y"' Q 'x v ^i.» u I w 18 called their stmpie value* Seven units, by this character, 7. itistheir value always when ijiight units, by this character, 8. emgle. Nine units, by this character, 9. Nine is the largest number which can be expressed by a single figure. There is another character, ; it is called a cipher, naught, or nothing, because it denotes the absence of a thing. Still it is of frequent use in expressing numbers. By these ten characters, variously combined, any number may be expressed. The unit 1 is but a single one, and in this sense it is called a unit of the^r^^ order. All numbers expressed by one fig- ure are units of the first order. ir 3» Ten has no appropriate character to represent it, but It is considered as forming a unit of a second or higher order, consisting of tens. It is represented by the same unit figure 1 as is a single thing, but it is written at the left hand of a cipher; thus, 10, ten. The fills the^r^^ place, at the right hand, which is the place of units, and the 1 the second place from the right hand, which is the place of tens. Being put in a new place, it has a new value, which is ten times its simple value, and this is what is called a local value. Questions. — IT !• What is a single thing called? What is a bumoer ? Give some examples. How many ways of expressing num- oers shorter than writing tnem out in words? What are they called? Which is the method in general use ? In the Arabic method, how many numbers have each a separate character? yi 2. How is one represented ? Make the characters to nine. What are these nine characters called ? Why ? What is the simple value of jgures? What is the largest number which can be represented by a single figure ? . What other character is frequently used ? Why is it called naught? How many are the Arabic characters? What are lumber;* expressing single things called ? IT 4. NOTATION AND NUMERATION. 11 There iftay be one, two, or more tens, just as there are one, two, or more units, or single things ; it takes ten cents to make one ten- cent piece ; just so it takes ten single things to make one ten. All figures in the second place express units of the 2d order, that is, units of tens. One ten and one unit, 11, are called eleven ; one ten and two units, 12, twelve, &c. In this way the units of the 1st order are united One ten is . Two tens are Three tens " Four tens Five tens Six tens Seven tens " Eight tens " Nine tens " cc (( • 10 ten. . 20 twenty. . 30 thirty. . 40 forty. . 60 fifty. • 60 sixty. . 70 seventy. . 80 eighty. . 90 ninety. One ten, one unit, 11 e.even. One ten, two units, 12 twelve. Note, Twenty, thirty, &c., are contractions for two tens, three tens, &c. with the tens^ that is, with the units of the 2d order, to form the numbers from 10 to 20, from 20 to 30, to 40, and so on to 99, which is the largest number that can be represented by two figures. The weeks in a year are 5 tens and 2 units, (5 of the sec- ond order and 2 of the first order now described,) and are expressed thus, 52, (fifty-two.) In the same manner express on your slate, or on the blackboard, the two orders united, so as to form all the numbers from 10 to 99. IT 4» Ten tens are called one hundred, which forms a unit of a still higher, or 3d order, and is ex- -i «• j pressed by writing two ciphers at the M-^ right hand of the unit 1, . . . thus, 100 one hundred. Note. When there are no units or tens, we ^00 two hundred, write ciphers in their places, which denote the 300 three hundred, absence of a thing, (^f 2.) &c. Qaestions. — If 3. How is ten represented ? What is it considered as lorming? Consisting of what? What place does the cipher fill?. The one? Where is unit's place, and where ten's place, counting from the right ? How much larger is the value of a figure in the place of tens than in the place of units ? In which place does it retain its sim pie value ? In ten's place, what is its value called? What is 1 ten and 1 unit called ? 1 ten and 2 units ? How are the numbers from 10 to 99 expressed? Of what is the number forty made up? Ans, 4 tens and no units. Sixty? What do you unite, to form the number twenty, three? tliirty-seven ? seventy-five? &c. Of what are twenty, thirty, &c., contractions ? What is the largest, and what the hast, numlier you ean express by one figure? by two figures? 12 NOTATION AND NUMERATION. IT 5, 6, Three hundred sixty-five, the days in a year, are expressed thus, 365 ; 3 being in the place of hundreds, 6 in the place of tens, and 5 in the place of units. After the same manner, the pupil may be required to unite the three orders, and express any number from 99 to 999. V S* We have seen that figures have ttoo values, viz., simple and local. The simple value of a figure is its value when standing alone ; thus, the simple value of 7 is seven. The local value of a figure is its value according to its dis' tancefrom the place of units ; thus, the local value of 7, in the number 75, is 7 tens, or seventy, while its simple value is seven ; in the number 756, its local value is sev^ hundred. Note. From the fact that 10 is 1 more than 9, it follows, as may be found by trial, that the local value of every figure at the left of units, except 9, exceeds a certain number of nines by the simple value of the fi^re. . Take the number 623 ; 2 (tens) is 2 more than 2 nines, and 6, (hundreds,) 6 more than a certain number of nines. On this principle is founded a method of proof in the subsequent rules, by casting out the nines. V ©• Ten hundred make one thousand, which is called a unit of the next higher, or 4th order, consisting of thousands^ and is expressed by writing three ciphers at the right hand of the unit 1, giving it a new local value ; thus, 1000, one thou- sand. * To thousands succeed tens and hundreds of thousands, forming units of the 5th and 6th orders. Questions. — IF 4, What are 10 tens called ? What do they form ? How many places are required to express hundreds ? How much does I cipher, placed at the right hand of 1, increase it ? 2 ciphers ? How do you express two hundred? &c. What are 4 hundreds, 9 tens, and 5 units called? How is one hundred ninety-three expressed? What place does the 3 occupy ? the 9 ? the 1 ? How do you express the ab- sence of an order ? How is the number of days in a year expressed ? ^ 5, How many values have figures ? What are they ? What is the simple value ? local value ? What is the value of 5 in 59 ? Is it its simple, or a local value ? Is the value of 8, in 874, simple or local ? of the 7? of the 4? 1 6. How do you express one thousand? seven thousand? A thou- sand is a unit of what order ? How many thousands are 30 hundreds ? What after thousands, and of what order ? The 6th order is what ? In writing nine hundre 1 and two thousand and nine, where do you plac6 ciphers ? Why ? T7,8. NOTATION AND NUMERATION. 13 TABIiB. T y. In this table of the six orders now described, you see the unit 1 moving from right to left, ^nd at each removal forming the unit of a higher order. There are other or- ders yet undescribed, to form which the imit '' moves onward still towards the left, its value being increased ten times by each removal. Note 1. The Ordinal numbers, 1st, 2d, 3d, &c., may be called indices of their respective orders. Note 2. VarioiLS Readings. In the number 546873, the left hand figure 5 expresses 5 units of the 6th order, or it may be rendered in the next 'ower order with the 4, and together they may be i a a n read 54 units of the 6th order, (ten thousands,) and 10 connecting with the 6, they may be read, 546 units 10 of the 4th order, or 546000. Hence, units of any 100000 higher order may be rendered in units of any lower f » f » f f ^^' 9 9 9 9 9 9 To hundreds of thousands succeed units, tens, and hunr dreds of millions. IT 8. To millions succeed billions, trillions, quadrillions, quintillions, sextillions, septillions, octillions, noniUions, decil- lions, undeciUions, duodecillions, tredecillions, &c., to each of which, as to units, to thousands, and to millions, are assigned three places, viz., units, tens, hundreds, as in the following ex- amples : Questions. — f 7« How is the unit 1 of the 1st order made a unit of the 2d order? of the 3d order, &c., to the 6th order ? What may the ordinal numbers, 1st, 2d, 3d, &c., be called ? 7 units of the 6th order are how many units of the 4th order ? The teacher will midtiply such questions. What is the least, and what the largest, number which can be expressed by 2 places? 3 places? &c. What after hundreds of thou- sands ? Of what order will millions be ? tens of millions ? hundreds of millions ? ^ 8« What after millions ? How many places are allotted to bil- lions? to trillions? &c. Give the names of the orders after trillions. In reading large numbers, A^hat is frequently done? Why? The Isi period at the right is the period of what ? the 2d? the 3a? the 4th? dec. 2 \A NOTATION AND NUHERATION. T 9. i i i t i i i '^ i ^ i ^ « •§ 1 i s =3 J -i 0) CD J^ ^ m fi ^ *^ %m %*^ Urn ^ ^ ^ o o o o o o o "S to CO 00 00 JD '^ "Vf "!? ''C 'w _§ 2 S S2 2 "2'2oon'S«5'5ooj5'5«3'2«fi'2aB5'5oo2'2«a & WEhP W^P MHP WHP WHP WHP WHt3 WHUs 3062715203 174592837463512 3, 8 2, 7 1 5, 2 3, 1 7 4, 5 9 2, 8 3 7, 4 6 3, 5 1 2 To facilitate the reading of large numbers, we may point them off into periods of three figures each, as in the 2d exam- ple. The names and the order of the periods being known, this division enables us to read numbers consisting of many figures as easily as we can re^f^ those of only three figures. Thus, in looking at the above examples, we find the first pe- riod at the left hand to contain one figure only, viz., 3. By looking under it, we see that it stands in the 9th period from units, which is the period of septillions ; therefore we read it 3 septillions, and so on, 82 sextillions, 715 quintillions, 203 quadrillions, 174 trillions, 592 billions, 837 millions, 463 thou- sands, 512. IT ©• From the foregoing we deduce the following princi- ples: — Numbers increase from right to left, and decrease from left to right, in a ten-fold proportion ; and it is A Fundamental Law of the Arabic Notation ; that, Questions. — ^9* How do numbers increase ? how deurease, and in what proportion ? To what is 1 ten equal ? 1 hundred? 1 thousand? &c. To what are 10 units equal? 10 hundreds? &c. What is a fun- damental law of the Arabic notation ? What is notation ? numera- tion? How do you write numbers? read numbers? If you were to write a number containing units, tens, hur ireds, and millions, but no thousands, how would you express it ? f 10. NOTATION AND NUMERATION. 15 I. Removing any figure one place towards the Ufty in* creases its value ten timesy and II. Removing anyfigure one place towards the right, de- creases its value ten times. The expressing of numbers as now shown is called Nota- tion. The reading of any number set down in figures is called Numeration. To write numbers, — Begin at the left hand, and write in their respective places the units of each order mentioned in the number. If any of the intermediate orders of units be omitted in the number mentioned, supply their respective places with ciphers. * To read numbers, — Point them otf into periods of three figures each, beginning at the right hand ; then, beginning at the left hand, read each period separately. Let the pupil write down and read the following numbers : Two million, eighty thousand, seven hundied and five. One hundred mHlion, one hundred thousand and one. Fifty-two million, sixty thousand, seven hundred and throe. One hundred thirty-two billion, twenty-seven million. Five trillion, sixty billion, twenty-seven million. Seven hundred tnllion, eighty-six bilUon, and nine. Twenty-six thousand, five hundred and fifty men. Two million, four hundred thousand dollars. Ninety-four billion, eighty thousand minutes. Sixty trillion, nine hundred thousand miles. Eighty-four quintillion, seven quadrillion, one hundred million grains of «and. IT lO. Numbers are employed to express quantity. Quantity^ is anything which can be measured. Thus, Time is quantity, as we can measure a portion of it by days, hours, &c. DistaTice is quantity, as it can be measured by miles, rods, &c. By the aid of numbers quantities may either be added to- gether, or one quantity may be taken from another. Arithmetic is the art of making calculations upon quanti- ties by means of numbers. Quel tion 8. — IF 10. Numbers are employed to express what? What is quantity ? B^ what is a quantity of grain measured ? a quan- tity of cloth*? What is arithmetic ? What is an abstract number ? a denominate number? What is the unit of a number 1 What is the unit value of 8 bushels ? of 16 yards ? of 20 pounds of sugar ? of 3 quarts of milk ? of 9 dozen Of buttons ? of 18 tons of hay f of 16 hogs* heads of molasses? 16 ADDITION OF SIMPLE NUMBERS. IT 11. A number applied to no kind of thing, as 6, 10, 18, 36, is called an abstract number. A number applied to some kind of thing, as 7 horses, 25 dollars, 2e50 men, is called a denominate number. The unity or unit value of a number, is one of the kind which the number expresses ; thus, the unit of 99 days is 1 day ; the unit of 7 dollars is 1 dollar ; the unit of 1:5 acres is 1 acre. In like manner the unit of 9 tens may be said to be 1 ten ; the unit of 8 hundred to be 1 hundred; the unit of 6 thousand to be 1 thousand, &c. ADDITION OF SIMPLE NUMBERS. IT 11* 1* James had 5 peaches, his mother gave him 3 more ; how many had he then ? Ans. 8. Why ? Ans. Because 5 and 3 are 8. 2. Henry, in one week, got 17 merit piarks for perfect les- sons, and 6 for good behavior; how many merit m^ks did he get ? Ans, . Why ? 3. Peter bought a wagon for 36 cents, and sold it so as to gain 9 cents ; how many cents did he get for it ? 4. Frank gave 15 walnuts to one boy, 8 to another, and had 7 left ; how many walnuts had he at first ? 5. A man bought a chaise for 54 dollars ; he expended 8 dollars in repairs, and then sold it so as to gain 5 dollars ; how many dollars did he get for the chaise ? The putting together of two or more numbers,' (as in the foregoing examples,) so as to make one whole nuniber^ is called Addition^ and the whole number is called the Sum^ or Amount. 6. One man owes me 5 dollars, another owes me 6 dol- lars, another 8 dollars, another 14 dollars, and another 3 dol- lars ; what is the sum due to me ? 7. What is the amount of 4, 3, 7, 2, 8, and 9 dollars ? 8. In a certain school, 9 study grammar, 15 study arith- metic, 20 attend to writing, and 12 study geography ; what is the whole number of scholars ? Questions. — ^ 11. What is addition? What is the answer, oi numl)er sought, called? What is the sign of addition? What does it show? How is it sometimes read? Whence the word plus^ and what Is its signification? What is the sign of equality, and what does it sbcw? Til. ADDITION OP SIMPLE MUMBBBS. 17 Signs. — A cross, -f-, one line horizontal #nd the other p.er- pendicular, is the sign of Addition. It shows that numbers ^th this sign between them are to be added together ; thus, 44-7 + 14+16 denote that 4, 7, 14, and 16 are to be added together. It is sometimes read plus^ which is a Latin word signifying 7nore, Two parallel, horizontal lines, =, are the sign of Equality, It signifies that the number before it is equal to the number after it ; thus, 5 + 3=s=8 is read 5 and 3 are 8; or, 6 plus 3 are equal to 8. In this planner let the pupil be instructed to commit the following ADDITION TABLJS. 2 + 0: 2+1 2 + 2 2+3: 2 + 4: 2 — 6 2--6 2 — 7 2 — 8 .2 + 9 2 3 4 5 6 7 8 9 10 11 6 + 0: 6— 1 6 + 2 6--3 6 + 4 6+5 6 + 6 6 + 7 6 + 8 6-^9 6 7 8 ft 9 10 11 12 13 14 15 3H [-0= 3 f 4H [-0— 4 5H 1-0= 5 3- -1— 4 4- -1= 5 5- -1= 6 3- -2— 5 4- -2= 6 5- -2— 7 3- -3= 6 4- [-3— 7 6- k3— : 8 3- -4— 7 4- -4= 8 5- -4= 9 3- -5= 8 4- h5— 9 5- -5 = 10 3- k6-r 9 4- -6 — 10 5- -6=11 3- -7—10 4- -7 = 11 5- -7— :12 3- f-8— 11 4- -8=12 5- -8=13 3- ^-9 — 12 4- -9—13 5- -9—14 7- [-0— 7 8- 1-0= 8 9H kO= 9 7- -1= 8 8- -1— 9 9- -1 = 10 7- -2= 9 8- -2—10 9- -2=11 7- -3=10 8- -3—11 9- -3—12 7- -4=11 8- -4=12 9- -4—13 7- -5—12 8- -5=13 9- -5=14 7- -6 — 13 8- -6 = 14 9- -6=15 7n -7 = 14 8- -7—15 9- -7=16 7-1 -8=15 8- -8=16 9- -8=17 7- -9=16 8- -9=17 9- -9 = 18 5 8 4 6 2 7 3 ■ 9 == how many ? ■ 7 = how many ? . 3 + 2 :=: how many ? . 4 - - 5 = how many ? . - - 4 + 6 = how many ? ■ l--0--8 = how many ? . - - 9 - -4= how many ? 2* 18 ADDITION OF SIMPLB ffUMBBIUi. TI2 -. :;;--()-. 4--D=s HOW many _-3--5--7--8 = bow many hi *J .1 W^ I A-k irk ^^m^tm «^ 9H h^H h^H ^44 1- -3- r6- [-7- 1- -2- k3- -'^H 8- -9- -0- -2- 6- -2- -5- -0- 5 = how many ? 8 = bow many ? 5 4- 6 = how many ? 4X5 = howmaDV? u -f- o == iiuw iiiiuiy I • 44-5 = how many ? • 8 -|- 3 = how many ? IT 19* When the numbers to be added are smalls the ad* dition is readily performed in the mivid^ and this is called mental arithmetic; but it will frequently be more convenient, and even necessary, when the numbers are large^ to write them down before adding them, and this is called written arithmetic. 1. Harry had 43 cents, his father gave him 25 cents more ; how many cents had he then ? Solution.— One of these nambers contains 4 tens and 3 units. The other number contains 3 tens and 5 units. To unite these two numbers together into one, write them down one un- der the other, placing the units of one number directly under units of the other, and the tens of one number directly under tens of the other, and draw a Une un- derneath. 43 cents, 25 cents. 43 cents. 25 cents. 8 43 cents. 25 cents. Ans. 68 cents. Beginning at the column of units, we add each column separately ; thus, 5 units and 3 units are 8 units, which we set down in units' place. • We then proceed to the column of tens, and say, 2 tens and 4 tens are 6 tens, or 60, which we set down directly under the column in tens' place, and the work lb done. It now appears that Harry's whole number of cents is 6 tens and 8 units, or 68 cents ; that is, 43 -f- 25 ;= 68. Units are written under units, tens under tens, &c. ; be- cause none but figures of the same unit value can be added to each other ; for 5 units and 3 tens will make neither 8 tens nor 8 units, just as 5 cows and 3 sheep will make neither 8 cows nor 8 sheep. Questions* — IT IS* What distinction do you make between men- tdl and written arithmetic ? How do you write numbers for addition ? Where do you begin to add ? and where do yoa set the amount ? How do yon proceed 7 Why do yoa write units under ui^ts, tens under tens, fiect m T 13. AINOITION €»* SnO^LB NUMBBIM. 19 2. A farmer bought a chaise tor 210 dollars, a horse fot 70 dollars, and a saddle for 9 dollars ; what was the whole amount ? Write the numbers as before directed, with units under units, tens under tens, 6cc. OPERATION. Chaise, 210 dollars* Horse, 70 doUart. . "^^''j^v: PS °i?^*" r^'*f.'«*'\' SaddU, 9 dollars. ;^«^8^^^ hundred. 2; that », 810+ Answer f 289 dollars. After the same manner are performed the following exam- ples, in which the amount of no column exceeds nine, 3. A man had 15 sheep in one pasture, 20 in another pas* ture, and 143 in another ; how many sheep had he in the three pastures ? 15 -|- 20 -[- 143 = how many ? 4. A man has three farms, one containing 500 acres, an- other 213 acres, and another 76 acres ; how many acres in the three farms ? 500 + 213 + 76 = how many ? 5. Bought a farm for 2316 dollars, and afterwards sold it so as to gain 550 dollars ; what did I sell the farm for ? 2316 -j- 550 = how many ? 6. A chair-maker sold, in one week, 30 Windsor chairs, 36 cottage, 102 fancy, and 21 Grecian chairs ; how many chairs did he sell ? 30 li- 36 + 102 + 21 = how many ? 7. A farmer, after selling 500 bushels of wheat to a com- mission merchant, 320 to a miller, and sowing 117 bushels, found he had 62 bushels left ; how many bushels had he at first? 500 + 320 + 117 + 62 = how many? 8. A dairyman carried to market at one time 231 pounds of butter, at another time 124, at another 302, at anoUier 20, and at another 12 ; how many pounds, did he carry in all ? Ans, 689 pounds. 9. A box contains 115 arithmetics, 240 grammars, 311 geogi;nphies, 200 reading books, and 133 spelling books ; how many books are there in the box ? Ans, 999. V 13. Hitherto the amount of any one column, when added up, has not exceeded 9, ai d consequently has been ex- pressed by a single figure. But it will frequently happen that the amount of a single column will exceed 9, requiring two or morejigiires to express it. 1. There are three bags of money. The first •ontains 876 JO ADDITION OP SIMPLE NtJMBEKS. f li JollaTs, the second 653 dollars, the third 426 dollars ; what is the amount contained in all the bags ? OPERATION. Solution. — Writing the numbers as First bag, 876 dollars, already described, we add the units, and Second " 653 " ^^^ them to be 15, equal to 5 units, which Third " 426 * ^^ write in units' place, adding the 1 ten with the tens ; which being added TTZZ together are 15 tens, equal to 5 tens, to 1955 " i>e written in tens' place, and 1 hundred, to be added to the hundreds. The hun- dreds being added are 19, equal to 9 hundreds, to be written in hun- dreds' place, and 1 thousand, to be written in thousands' place. Ans, 1955 dollars. Proof. — We may reverse the order, and, beginning at the top, add the figures downwards. If the two results are alike, the wmrk may be supposed to be right, for it is not likely that the same mistake wiU be made twice, when the figures are added in a different order. Note. — Proof by the excess of nines. If the work be right, thefs will be just as many of any small number, as 9, with the same re- mainder, in the amount, as in the several numbers taken together. Hence, OPERATION ^^ *^® upper number, 8 (hundreds) is 8 more than a * certain number of nines, {% 5) 7 (tens) is 7 more. Adding the 8 and 7, and the 6 units together, the sum is 21 = 2 nines and 3 remainder, which we set down at the right hand, as the excess of nines in this number. In the same manner, 5 is found to be the excess of nines in the second number, and 3 in the tliird number. These several excesses being added together, make 1 nine and an excess of 2, which is the same as the excess of nines in the general amount, found in the same manner. This method will detect every mistake, except it be 9, or an exact number of nines. To find what will be the excess .after casting the nines out of any number, begin at the left hand, and add together the figures which express tbe number; thus, to cast the nines out of 892, we say 8 (passing over 9) -}-2 (dropping 9 from the sum) =* 1. From the examples and illustrations now given, we derive the following I. Write the numbers to be added, one under another, plac- 876 3 653 5 426 3 1955 2 Questions. — If 13. If the amount of the column does not exceed 9, what do you do ? What whe». it exceeds 9 ? How do you add each column? What do you do wita the amount of the left column? For what number do you carry ? If the amount of a column be 36, what would you set down, and how many would you carry ? On what prin* cjple do you do this ? How is addition proved ? Why ? Repeat the rule for aiddition. H 13. ADDITION OF SIMPLE NUMBEBS. 21 ing units under units, tens under tens, &c., and draw a line underneath. II. Begin at the unit column, and add together all the fig- ures contained in it ; if the amount does not exceed 9, write it under the column ; hut if it exceed 9, write the units in units* place, and cjurry the tens to the column of tens. III. Add each succeeding column in the same manner, and set down the whole amount of the last column. BXAMPUBS FOR PRACTICS. 286370542106 1 4367583021463 3 1074293 15638 1752349713620 6253034792 6081275306217 247135 5652174630 128 8673 87032634720 13 4. Add together 587, 9658, 67, 431, 28670, 85, 100000, 6300, and 1. Amount, 145799. 5. What is the amount of 8635, 7, 2194, 16, 7^1. 93, 5063, 135, 2196, 89, and 1225? Am, 27074. 6. A man heing asked his age, answered that he left Eng- land when he was 12 years old, and that he had afterwards spent 5 years in Holland, 17 years in Germany, 9 years in France, whence he sailed for the United States in the year 1827, where he had lived 22 years ; what was his age ? Ans. 65 years. 7. A company* con tract to build six warehouses ; for the first "they receive 36850 dolls. ; for the second, 43476 dolls. ; for the third, 18964 dolls. ; for the fourth, 62840 dolls. ; for the fifth, 71500 dolls. ; for the sixth, as much as for the first three ; to what do these contracts amount? -Atw. 332920 dollars. 8. James had 7 marbles, Peter had 4 niarbles more than James, and John had 5 more than Peter ; how many marbles in all ? Ans. 34. 0- 9. There are seven men ; the first man is worth 67850 dol- 4ars ; the second man is worth 2500 dolls, more than the first man ; the third, 3168 dolls, more than the second ; the fourth, 16973 dolls, more than the third ; the fifth, 40600 dolls, more than the fourth ; the sixth, 19888 dolls, more than the fifth ; and the seventh, 49676 dolls, more than the sixth ; how many dollars are they all worth ? Ans, 784934 dollars. 10. What is the interval in years between a transaction 22 ADDITION OF SIMPLE NUMBERH. H 14 that happened 275 years ago, and one that will happen 125 years hence ? Ans, 400 years. 11. What is the amount of 46723, 6742, and 986 dollars ? 12. A man has three orchards ; in the first there are 140 trees that bear apples, and 64 trees that bear cherries ; in the second, 234 trees bear apples, and 73 bear cherries ; in the third, 47 trees bear plums, 36 bear pears, and 25 bear cher- ries ; how many trees in all the orchards, and how many of each kind ? Ans. 619 trees ; 374 bear apples ; 162 cherries ; 47 plums; and 36 pears. . 13. A gentleman purchased a farm for 7854 dollars ; he paid 194 dollars for having it drained and fenced, and 300 dollars for having a bam built upon it ; how much did it cost him, and for how much must he sell it, to gain 273 dollars ? M { It cost him 8348 dollars. Ans. < He must sell it for 8621 dollars. IT 14. Review of Numeration and Addition. l^nestions. — What are numbers? What are the methods of expressing numbers? What is numeration? notation? fundamental law in the Arabic notation ? How does the Arabic differ from jhe Ro- man method? What is understood by units of different orders ? What is quantity ? Arithmetic ? What is understood by the simple value of figures? the local value? the unit value of a number? Explain the difference between an abstract and a denominate number. What is addition ? the rule ? proof? For what number do you carpy, and why ? EXERCISES. 1. Washington was bbrn in the year of our Lord 1732 ; he was 67 years old when he died ; in what year did he die ? Am, 1799. 2. The invasion of Greece by Xerxes took place 481 years before Christ ; how long ago is that this current year ? 3. There are two numbers ; the less is 8671, the difference between the numbers is 597 ; what is the greater number ? Ans. 9268. 4. A man borrowed a sum of money, and paid in part 684 dollars ; the sum left unpaid was 876 dollars ; what was the sum borrowed ? 5. There are four numbers ; the first 317, the second 812 the third 1350, and the fourth as much as the other three ; what is the sum of them all ? Ans. 4958. 6. A gentleman left his daughter 16 thousand 16 hundred t X T 15. SUBTRACTION OF SIMPLE NUMBSBd. 23 and 16 dollars ; be left his son 1800 more than his daughter; what was his son's portion, and what was the amount of the whoU estate ? a S Son's portion, 19416. ^ns. I ^yj^^j^ ^g^^^^ 37Q32 7. A man, at his death, left his estate to his four children, who, after paying debts to the amount of 1476 dollars, re- ceived 4768 dollars each ; how much was the whole estate ? Ans, 20548. 8. A man bought four hogs, each weighing 375 pounds ; how much did they all weigh ? Ans. 1500. -^ 9. The fore quarters of an ox weigh one hundred and eight pounds each, the hind quarters weigh one hundred and twenty-four pounds each, the ^hide seventy^six pou^fis, and the tallow sixty pounds ; what is the whole weight of the ox ? Ans, 600. 10. The imports into the several States in 1842 were as follows : Me. 606864 dollars, N. H. 60481, Vt. 209868, Mass. 17986433, R. I. 323692, Ct. 335707, N. Y. 57875604, N. J. 145, Pa. 7385858, Del. 3557, Md. 4417078, D. C. 29056, Va. 316705, N. C. 187404, S. C. 1359465, Ga. 341764, Al. 363871, La. 8033590, O. 13051, Ky. 17306, Tenn. 5687, Mich. 80784, Mo. 31137, Fa. J76980 dollars ; what was the entire amount ? Am. 100162087. SUBTRACTION OF SIMPLE NUMBERS. IT IS* 1. Charles, having 18 cents, bought a book, for which he gave 6 cents ; how many cents had he left ? 2. John had 12 apples ; he gave 5 of them to his brother ; how many had he left ? 3. Peter played at marbles ; he had ^3 when he began, but when he had done he had only 12 ; how many did he lose ? 4. A man bought a cow for 17 dollars, and sold her again for 22 dollars ; how many dollars did he gain ? 5. Charles is 9 years old, and Andrew is 13 ; what is the difference in their ages ? 6. A man borrowed 50 dollars, and paid all bui 18 ; how many dollars did lie pay ? that is, take 18 from 50, and how many would there be left ? The taking of a less number from a greater (as in the fore- going examples) is called Subtraction, The greater number 24 SUBTRACTION OP SIMPLE NUMBERS. Tie. is called the Minttend, the less number the Subtrahend^ and what is left after subtraction is called the Difference^ or Ri mainder, 7. If the minuend be 8, and the subtrahend 3, what is the difference or remainder ? Ans» 5. 8. If the subtrahend be 4, and the minuend 16, what is the remainder ? Sign. — A short horizoatal line, — , is the sign of subtrac- tidn. It is usually read minus, which is a Latin word signi- fying less. It shows that the number after it is to be taken from the number before it. Thus, 8 — 3=*= 5 is read 8 minus or less 3 is equal to 5 ; or, 3 from 8 leaves 5. The latter expression is to be used by the pupil in committing the following . SUBTRACTION TABLE. 2 3 4 5 6 7 8 9 10 2 2 2 2 2 2 2 2 2 a 1 2 3 4 5 6 7 •8 3 \ 5 -3: — 3: — 3: 7 — 3 9 — 4 12 — 3: 19—4: 1 2 how many ? how many ? how many ? how many ? how many ? 18 22 33 41 ■ 7 = how many ? . 7= how many? 13 = how many ? — 5= how many? — 15 = how many ? ^10. When the numbers are small, as in the foregoing examples, the taking of a less number from a greater is readily done in the mind ; but when the numbers are large, Qnestions* — IF 15. What is subtraction? Wliat is the greater number called? the less number? that which is left after subtraction? Wliai is the sign of subtraction ? How is it usually read? What dees minus mean ? What does the sign of subtraction show? H 16. SUBTRACTION OP SIMPLE NUMBERa 25 the operation is most easily performed part at a time, pnd therefore it is necessary to tvrite the numbers down before performing the operation. 1. A farmer, having a flock of 237 sheep, lost 114 of them by disease ; how many had he left ? Here we have 4 units to be taken from 7 units, 1 ten to li© taken from 3 tens, and 1 hundred to be taken from 2 hundreds. It will therefore be most convenient to write the less number under the greater, observing, as in addition, to place units under units, tens under tens, &c., thus : OPERATION. Solution. — We begin with the From 237 the minuend, units, saying, 4 (units) from 7, (units, ) Take lU the subtra/iend *°^ ^^®^® ^^^^^^ ^' (units,) which we set down directly under the column , in units' place. Then proceeding to 123 the re7nat7ider. the next column, we say, 1 (ten) from 3, (tens,) and there remain 2, (tens,) which we set down in tens^ place. Proceeding to the next column, we say, 1 (hundred) from 2, (hundreds,) and there remains 1, (hun- dred,) which we set down in hundreds^ ulace, and the work is done. It now appears that the number of sheep left was 123 ; that is, 237 — 114=123, Aw5. Note. — We write units under units, tens under tens, &c., that those of the same unit value may be subtracted from each other ; for we can no more take 3 tens from 7 units than we can take 3 oows from 7 sheep. Examples in which each figure in the subtrahend is less than the figure aJ)ove it, 2. There are two farms ; one is valued at 3750, and the other at 1500 dollars ; what is the difference in the value of the two farms ? . ' Ans. 2250. 3. A man's property is worth 8560 dollars, but he has debts to the amount of 3500 dollars ; what will remain after paying his debts ? Ans, 5060. 4. From 746 subtract 435. Rein, 311. 5. From 4983 subtract 2351. Em. 2632. 6. From 658495 subtract 336244. Rem, 322251. ' 7. Ffom 8764292 subtract 7653181. Rem. 1111111. Qnestions. — T !€• When the numbers are small, hew may the ^ration be performed ? "When they are large, what is. more conve- ^nt ? How are the two numbers to be written ? Where do you begin 4it fubtractioQ 7 8 26 gUBTRACTION OF SJMPLE NUMBEB3. H 17. IT IT. 1. James, having 15 cents, bought a pen-knife, for which he gave 7 cents ; how many cents had he left ? OPERATION. 15 cents. A difficulty presents itself here ; for we 'Cannot 7 cents, take 7 from 5 ; but we can take 7 firom 15, and — . there will remain 8. 8 cents left. 2. A man bought a horse for 85 dollars, and a cow for 27 dollars ; what did the horse cost him more than the cow ? OPERATION. Solution . — The same difficulty pre»ents itself horn g5 ' as in the last example, that is, the unit figure in the 2*9 subtrahend is greater than the unit figure in the minu- end. To obviate this difficulty, we may take 1 (ten) from the 8 (tens) in the minuend, which will leave 7 (tens,) and add it to the 5 units, making 15 units, (7 tens -|- 15 units SB 85,) thus, TENS. UNITS. . 7 15 W© ^^^ ^^e 7 units from 15 unitt, and 2 tens from 277 tens, and have 5 tens and 8 units, or 58 remainder ; that is 85 — 27 =s 59 dollars more for the horse than - t^ for the cow. 08 . The operation may be shortened as follows : OPERATION. ^^ ^^^^ ^ ^'^ ^^^ ^ "^^^ ^^ ^^^ minuend, Hiww 8/5 dollars ^"^^ ^ ^"^^ ^'^^ ^ units in the subtrahend. We "^ 00 aouars. ^^^ nov,-^ in ike mind, suppose 1 ten taken from LoWf 27 " ^Q Q tgng^ which would leave 7 tens, and this — 1 ten we can suppose joined to the 5 units, Diff' 58 ** making 15. We can now take 7 from 15, as before, and there will remain 8, which we set down. The taking of 1 ten out of 8 tens, and joining it with the 6 units, is called borrowing ten. Proceeding to the next higher order, or tens, we must consider the upper figure, 8, from which we bor- rowed, 1 less, calling it 7 ; then, taking 2 (tens) from 7, (tens,) there will remain 5, (tens,) which we set down, making the difference 58 dollars, Ans. Qaestions. — ^ IT. In subtracting 7 from 15, what difficulty pre- sents itself ? How do you obviate it ? In taking 27 from 85, instead of taking 7 from 5 what do you take it from ? Whence the 15 ? From what do you subtract she 2 tens ? Why not from 8 tens instead of 7 tens? What is this operation called? Expjain how the operation is performed in example 3. There is another method, often practised, erroneously called borrowing teuj — explain the principle on which it is done. Wien we subtract units from units, of what name will ths remainder be? tens from tens, what? hundreds from hundreds, what? f 1& SUJiTRAOTiON 0^ mmiM NUlfBEHa 27 Non, -^ It kas bean usoal to perform sobtractioD, where the figure in the subtrahend is larger than the figure above it, on another prin- ciple. If to two unequal numbers the same number be added, the dif- ference between them will remain the same. Thus, the difference between 17 and 8 is 9, and the d^erenee between 27 and 18, eaeh being increased by 10, is also 0. Take the last example. g * y^ ' Adding 10 units to 5 units in the minuend, and 1 ^ J ten to 2 tens in the subtrahend, we have increased _^____^__ • both by the same number, and the remainder is not g g altered, being 58. This method, which has been erroneously called borrowing ten, may be practised by those who prefer, though the former is more ample and equally convenient. 3. From 10000 subtract 9« OFERATioN. SoLTTTioar. •— lu. thjs csxaaiple we have units {rom 10000 * ^^^^h ^ subtract 9 units, and going to tens of* the Q ^ minuend, we have tens, nor huiNkeds, nor thousands ; but we have 1 ten thousand 'from which, borrowing 10 units, we have 9990, that is, 9 thousands, 9 hundreds 9991 and 9 tens left. Taking 9 units from 10 units, we have 1 unit, then no tens in the subtrahend from 9 tens m the minuend leave 9 tens, no huiulreds from 9 hundreds leave 9 hundreds, no thousands firom 9 thousands leave 9 thousands, 4. A man borrowed 713 dollars and paid 475 dollars ; how much did he then owe ? Ans, 238 dollars. 5. From 1402003 take 681404. Rem. 720599. 6. What is the difference between 36070324301 and 280- 40373315 ? Ans, 8029950986. 7. From 81324036521 take 2546057867. Bern. 78777978654. T 18. To PKOVE Addition and Sttbtraction. — Addition and subtraction are the reverse of each other. Addition is putting together ; subtraction is taking asunder. 1. A man bought 40 sheep 2. A man sold 18 sheep and sold 18 of them ; how and had 22 left ; how many many had he left ? ha4 he at first ? 40 — 18 = 22 sheep left. 18 -f 22 = 40 sheep at first. Am. Ans. Hence, subtraction may be proved by addition, and addition by subtraction. To prove sttbtraction, add the remainder to the subtrahend, and, if the work is right, the amount will be equal to the fninuend. 28 SUBTRACTION OP SIMPLE NUMBERS. If 18 To prove addition^ subtract, successively, from the amount, the several numbers which were added to produce it, and, if the work is right, there will be no remainder, • Thus 7 -j- 8 4-6 = 21; ^oo/, 21 — 6=15, and 15 — 8 = 7, and 7 — 7 = 0. Note. — Proof by excess ofrnnes. We may cast ont the nines in the remainder and subtrahend ; if the excess equals the excess found by casting out the nines from the minuend, the work is presumed to be right. From the remarks and illustrations now given, we deduce the following RUIiE. L Write down the numbers, the less under the greater, placing units under units, tens under tens, &;c., and draw a line under them. II. Beginning with units, take successively each figure in the lower number from the figure over it, and write the re- mainder directly below. III. When a figure of the subtrahend exceeds the figure of the minuend over it, borrow 1 from the next left hand figure of the minuend ; and add it to this upper figure as 10, in which case the left hand figure of the minuend must be considered one less. Note. — Or when the lower figure is greater than the one above it we may add 10 to the upper figure, and 1 to the next lower figure. BXAMPIiES FOR PRACTICI3. 1. If a farm and the buildings on it be valued at 10000, and the buildings alone be valued at 4567 dollars, what is the value of the land ? Ans, 5433 dollars. 2. The population of New York in 1830 was 1,918,608; in 1840 it was 2,428,921; what was the increase in ten years ? Ans. 510,313. 3. George Washington was bom in the year 1732, and died in the year 1799 ; to what age did he live ? Ans, 67 years. 4. The Declaration of Independence was published July 4th, 1776 ; how many years to July 4th the present year ? Qaestions. — IflS. Addition is the reverse of what ? Subtrac tion, of what ? How will you show that they are so ? How do yoo prove subtraction ? How can you prove addition by subtraction? B> peat the rule for subtraction. ' % 19. SUBTRACTION OF SIMPLE NUMBERS. 29 6. The Rocky Mountains, in N. A., are 12,500 feet above the level of the ocean, and the Andes, in S. A., are 21,440 feet ; how many feet higher are the Andes than the Rocky Mountains ? Ans, 8,940 feet. Note. — Let the pupQ be required to prove the following examples. 6. What is the difference between 7,648,203 and 928,671 ? Ans, 6,719,532. 7. How much must you add to 358,642 to make 1,487,945? A^is, 1,129,303. 8. A man bought an estate for 13,682 dollars, and sold it again for 15,293 dollars ; did he gain or lose by it ? and how much? Arts, 1,611 dollars. 9. From 364,710,825,193 take 27,940,386,574. 10. From a3 1,025,403,270 take 651,308,604,782. 11. From 127,368,047,216,843 take 978,654,827,352. IT 19. Review of Subtraction. Qnesiions. — What is subtraction? What is the rule? What is understood by borrowing ten ? Of what is subtraction the reverse ? How is subtraction proved ? How is addition proved by subtraction' « EXSlRCISES. 1. How long from the discovery of America by Columbus, in 1492, to this present year ? 2. Supposing a man to have been born in the year 1773, how old was he in 1847 ? Am. 74. 3. Supposing a.man to have been 80 years old in the year 1846, in what year was he born ? Ans, 1766. 4. There are two iiumbers, whose difference is 8764; the greater number is 1JC37 ; I demand the less. Aiis, 6923. 5. What number is that which, taken from 3794, leaves 865 ? Ans, 2929. 6. What number is that to which if you add 789, it will become 6350 ? Aiis. 5561. 7. A man possessing an estate of twelve thousand dollars, gave two thousand five hundred dollars to each of his two daughters, and the remainder to his son ; what was his son's share ? A7is, 7000 dollars. 8. From seventeen million take fifty-six thousand, and what will remain ? Ans. 16,944,000. 3^ 30 SUBTRACTION OP SIMPLE NUMBERS. 1 IJI 9. What number, together with these three, tiz., 1301, 2561, and 3120, will make ten thousand ? Am, 301S. 10. A man bought a horse for one hundred and fourteen dollars, and a chaise for one hundred and eighty-seven dol- lar< ; how much more did he give for the chaise than for the norse? 11. A man borrows 7 ten ddlar bills and 3 one dollar bills, and pays at one time 4 ten dollar bills and 5 one dollar bills ; how many ten dollar bills and one dollar bills must he aftei- wards pay to 'Cancel the debt? Ans, 2 ten doll, bills and 8 one doll. 12. The greater of two numbers is 24, and the less is 16 ; what is their diflference ? 13. The greater of two numbers is 24, and their difTerence 8 ; what is the less number ? 14. The sum of two numbers is 40, the less is 16 ; what is the greater ? JEXERCISBS IN ADDITION AND SUBTRACTION. 15. A man carried his produce to market ; he sold his pork for 45 dollars, his cheese for 38 dollars, and his butter for 29 dollars ; he received, in pay, salt to the value of 17 dollars, 10 dollars' worth of sugar, 5 dollars' worth of molasses, and the res4 in money ; how much money did he receive ? Ans, 80 dollars. 16. A boy bought a sled for 28 cents, and gave 14 cents to have it repaired ; he sold it for 40 cents ; did he gain or lose by the bargain ? and how much ? Ans, He lost 2 cents. 17. One man travels 67 miles in a day, another man fol- lows at the rate of 42 miles a day ; if they both start from tlie same place at the same time, how far will they be apart at the close of the first day ? of the second ? of the third ? of the fourth ? Ans, To the last, 100 miles. 18. One man starts firom Boston Monday morning, and travels at the rate of 40 miles a day ; another starts from the «arae place Tuesday morning, and follows on at the rate of 70 miles a day ; how far are they apart Tuesday night ? Ans, 10 miles. 19. A man, owing 379 dollars, paid at one time 47 dollars Ht another time 84 dollars, at another time 23 dollars, and at another time 143 dollars ; how much did he then owe ? Ans. 82 dollars. 20. Four men bought a lot of land for 482 dollars ; the first man paid 274 dollars, the second man 194 dollars less than 1 20. MULTIPUOATION OP SIMPLE NUMBERa 31 the first, and the third man 20 dollars less than the second ; how much did the second, the third, and the fourth man pay ^ ( The second paid ^, Ans. { The third pajd 60. ( The fourth paid 68. 21 Four men hought a horse ; the first man paid 21 dol- lars, the second 18 dollars, the third 13 dollars, and the fourth AS much as the other three, wanting 16 dollars ; how much did the fourth man pay ? and what did the horse cost ? Ans. Fourth man paid — dolls. ; horse cost 88 dolls. 22. From 1,000,000 take 1, and what remains? (See IT 17 iEr. 3.) MULTIPLICATION OF SIBIPLE NUMBERS. IT 90* 1. If one orange costs 5 cents, how many cents must I give for 2 oranges ? how many cents for 3 Granges ? for 4 oranges ? 2. One bushel of apples cost 20 cents ; how many cents must I give for 2 bushels ? for 3 bushels ? 3. One gallon contains 4 quarts ; how many quarts in 2 gallons ? in 3 gallons ? in 4 gallons ? 4. Three men bought a horse ; each man paid 28 dollars for his share ; how many dollars did the horse cost them ? 5. In one dollar there ax e one hundred cents ; how many cents in 5 dollars ? 6. How much will 4 pairs of shoes cost at 2 dollars a pair ? 7. How much will two poimds of tea cost at 43 cents a pound? 8. There are 24 hours in one day ; how many hours in 2 days ? in 3 days ? in 4 days ? in 7 days ? 9. Six boys met a beggar, and gave him 15 cents each ; how many cents did the beggar receive ? In this example we have 15 cents (the number which each boy gave the beggar) to be repeated 6 times, (as many times as there were boys.) When questions occur where the same number is to be repeated several times, the operation may be shortened by a rule called Mtdtipliaation, In multiplication the number to be repeated is called the Multiplicand, The number which shows how many times the multiplicand «s to be repeated, is called the Multiplier » S3 BfXTLTlPLICATIOIf OF 8IMPLB fWJdNSa, H 20 The result, or answer, is called the Product The multiplicand and multiplier taken together are called Factors, or producers, because when multiplied together they produce- the product. 10. There is an orchard in which are 5 rows of trees, and 27 trees in each row ; how many trees in the orchard ? In the first row ^ . . . 2^ trees. Solution.— The whole num- " second, • • • • 27 ** her of trees will be equal to the M fJiiffl^ * * 27 •* amount of five 27*8 added to- " -5"/?!' ^•* '• * '• '• i? « ""in^adding. we find that 7 Jijcny • • • • . ^f taken five times amounts to 35. We write down the five units. In the whole orchard, 135 trees, and reserve the three tens ; the amount of 2 taken five times is 10, and the 3, which we reserved, makes 13, which, written -at the left of units, makes the whole number of trees 135. If we have learned that 7 taken 5 times amounts to 35, and that 2 taken 5 times amounts to 10, it is plain we need write the number 27 but OTUXj and then, settipg the multiplier under it, we may say, 5 times 7 are 35, writing Multiplicand, 27 trees in eojch row. down the 5, and reserving Multiplier, . 5 rows. ^^^ 3 (tens) as in addition. Again, 5 times 2 (tens) are product, . 135 trees, Ans. i.1icW .Sfrvtd/'S 13, (tens,) as before. From the above example, it appears that multiplication is a short way of performing many additions, and it may be defined, — The method of repeating one of two numbers as many times as tliere are units in the other. Sign: — Two short lines, crossing each other in the form of the letter X, are the sign of multiplication. When placed between numbers it shows that they are to be multiplied together ; thus, 3 X 4 = 12, signifies that 3 times 4 are equal to 12, or 4 times 3 are equal to 12 ; and thus, 4 X 2 X 7 = 5Q, signifies that 4 multiplied by 2, and this product by 7, equals 56. Questions, — If 20. When questions occur in which the same number is to be repeated several times, how may the operation be short- ened ? In multiplication, what is the mnltiplicand? the multiplier? the product? What are factors ? Why? What is multiplication? Illus- trate by the two methods ofjperforming the 10th example. How do you define multiplication ? Waat is the sign 7 Repeat the table. 12D. BIULTIPUOATION OP SIBfPLE NUMBER& 83 Note. — Before any progress c^ be made in this rule, the follow iDg table must be committed perfectly to memory. MUL.TIPL.ICATIOX TABLE. 10 X 3 — 30 10 X 4 — 40 10 X 5 — 50 10 X 6 — GO 10 X 7:-r=r 70 10 X S — 80 10 X 9 = 90 10 X 10 = 100 10 X 11 — 110 10X12 — 120 11 X = iix 1 = 11 11 X 2 — 22 11 X 3 — 33 iix 4— 44 11 X 5 = 56 11 X 6 — 66 11 X 7 — 77 11 X 8— 88 11 X 9 — 99 11 X lo- 110 ll X 11 = 121 11 X IS- 132 IS X = 12 X 1 — 12 12 X 2 — 24 12 X 3 — 36 12 X 4 — 48 12 X 5 — 60 12 X 6 — 72 12 X 7r^ 84 12 X 8 — 96 12 X 9 — 108 12XlO=r 120 12 X Il- 132 ls X 12= 144 1^ BIULTIPLICATION OP SIMPLE NtJMB^tA. T21 9 X 2 = how many? 4x3 k2 = 24. 4x6 = how many ? 3x2x5 = how many « 8 X 9=:howmany? 7x1 X2 = howmany? 3 X 7=how many? 8x3 X 2=: how many? 5 X 5 = how many? 3 X 2 X 4 X 5s=ahow many DIAGRAM OF S TARS. # # # m # # « m mm* m ? irStl. 1. There are on a hoard, 3 rows of stars, and 4 stars in a row ; how many stars on the board ? A slight inspection of the diagram will show that the number of stars may be found by considering that there are either 3 rows of 4 stars each, (3 times 4 are 12,) or 4 rows of 3 stars each, (4 times 3-are 12;) therefore, we may use either of the given numbers for a multiplier, as best suits our convenience. We generally write the numbers as in subtraction, the larg^er uppermost, with units under units, tens under tens, &;c. Thus, Multiplicand, 4 stars. Multiplier y 3 rows. Note. — 4 and 3 are the faetorSf • — which produce the product 13. Product, 12 stars, Ans, This diagram of stars is commended to the particular at- tention of the pupil, as it is intended to make use of it here- after in illustrating operations in multiplication and also in division. First, you will notice the terms of the diagram, and their application. TERMS OF THE DIAGRAM, ( Using this term as a representation or Stars in a row. < symbol of the multiplicand and one factor ( of the product. fj j^ f J Using this tenn as a symbol of the mul- ^ ' \ tiplier and the other factor of the product. Using this term as a symbol of the pro^ duct, for when the stars in a row are taken as many times as there are rows of stars, then the product will be the whole number ^ of stars contained in the diagram. As the stars in a row symbolize the multiplicand, it follows that the multiplier (number of rows) in reality simply express- es the number of times the multiplicand (stars in a row) is to Number of stars. T21. IflJLTIPLICAnOlf OP miPLB NUMBEAa 86 be taken. Hence, to multiply by 1, (1 row of stars,) is to take the multiplicand (stars in a row) 1 time ; to multiply by 2, (2 TOWS,) is to take the multiplicand (stcLrs in a row) 2 times; to multiply by 3, (3 rows,) is to take the multiplicand (stars in a row) 3 times, and so on. Illustration. — What cost 7 yards of cloth at 3 dollars a yard? (7 rows, 3 stars in a row.) The two numbers as given in the question are both denominate; but how are they to be considered in the operation f The price of 7 yards will evidently be 7 times the price of 1 yard, that is, 7 times 3 dollars ; dollars (number of starsV is the thing sought by the question ; and hence, 3 dollars being of the same name as the thing or answer sought, is the true multiplicand. That num- ber which was yards in putting the question, being taken for the multiplier^ in this relation is not to be considered yards, but times of taking the multiplicand ; and hence, in the oper- ation, it must always be considered an abstract number For multiplication is a short way of performing many ad- ditions, and to talk of adding 3 dollars to itself 7 yards times is nonsense. But we can repeat 3 dollars as many tunes as 1 yard is repeated to make 7 yards. There is then a true multiplicand and a true multiplier. The true multiplicand is that number which is of the same name as the answer sought ; the true multiplier is that num- ber which indicates the times the true multiplicand is to be repeated, or taken ; but as it respects the operation^ it has been shown above that we may use either of the given num- bers as the multiplier ; that is, the multiplicand and multiplier may change places ; still, the product will always be of the same name as the true multiplicand. This application of the terms of the diagram to the terms of the question we shall call symbolizing the question. Questions. — 1 21. You have in your book a diagram of stars j «rhat is the first use made of it ? What is the difference between 4 times 7, and 7 times 4? Which of the given numbers may be used for the multiplier ? What are the terms of the diagram ? What do these terms ^mbolize ? 6 times 7 are 42, —which of these numbers is the multipli- cand? the multiplier? the product? and what, in the diagram, is a sym- bol of each ? What does the multiplier express ? Show by the diamm what it is to multiply by 1, by 2, by 3, &c. What must the multmlier always be considered? What do you understand by the true multipli- cand? by the true multiplier? What will the product always be ? Give an example to show that you understand these principles. What do you understand by symbolizing a question 7 30 MXTLTIPUOATION OP 80fPL£ NUHBEBA IT 29. NoTS. ^ Let the teacher see to it Chat these principles axe weO understood by the pupil before he proceeds. As the pupil advances, the teacher should, from time to time, refer him back to a review of these principles. f 33. 1. What will 84 barrels of flour cost, at 7 dollars a barrel ? Solution. — The price of 84 barrels will evidently be 84 times the price of 1 barrel, 7 stars in one row X by 84, number of rows, 7 dol- lars is the true multiplicand, &c. ; but as it will be more convenient, the multiphcand and multiplier may change places, and we may consider it 7 rows of 84 stars in a row, and multiply the number of barrels, 84, by the price of 1 barrel, tims — Writing the larger number uppermost, OPERATION. 8S in subtraction, (^J 16,) and tne multi- Multiplicand, 84 P^^^^ ""^^^ ^"^'^^^ ^^ ^^^® multiplicand, we muuipner,^ / ^^^.^^ ^ 28 (units) = 2 tens and 8 units ; T> J /ZQQ J 77 ^® ^^ down the & units in units' place, as irroductf oS8 dolls, \n addition, and reserving the 2 tens, we say, 7X8 (tens) = 56 (tens,) and 2 (tens) which we reserved, make 58 (tens,) or five hundred and 8 tens, which we set down at the left of the 8 units, and tlie whole make 588 dol- lars, the cost of 84 barrels of flour, at 7 dollars a barrel, Ans. 2. A merchant bought 273 hats, (stars in a row,) at 8 del lars each, (number of rows ;) what did they cost (number of stars) ? Ans. 2184 dollars. ' 3. How many inches are there in 253 feet, (stars in a row,) every foot being 12 inches (number of rows) ? • ^^^ o^ro^^'. Solution.— The product of 12, with each of the ^oo significant figures or digits, haviiM^ been committed to 12 memory from the multiplication. , :,b].', it is just as easy to multiply by 12 as by a single figure. Thus, 12 Ans, 3036 times 3 are 30, &c. 4. What will 476 barreb of fish cost, at 11 dollars a bar- rel ? Am, 5236 dollars. From these examples we deduce the following Qaestions. — If 22. How will yoo explain the first exa»nple? When you multiply units by units, what is your product ? When tens by units, what ? How can you multiply by 12 ? Ho\fr do you write down numbers for multiplication ? How do you perform multiplication when the multipU* does not exceed 12 ? t23. MITLTIlUGATIOIf (»* SIMPLB m)Masa& 37 SLTHLSU I. To set doion numbers for multiplication. Write down the multiplicand, under which write the multiplier, setting units under units, tens under tens, &c. II. To perform multiplication when the multiplier does not exceed 12. Begin at the right hand, and multiply each figure in the multiplicand by the multiplier, setting down and car- rying as in addition. KXAMPIiES FOR PRACTICE. 5. A farmer sold 29 bags of wheat, each bag containing 3 bushels ; how many bushels did he sell ? 29 X 3 :=: how many? 6. A farmer, who had two farms, raised 361 bushels of wheat on one, and 5 times as much on the other ; how many bushels did he raise on both ? An^. 2166 bushels. 7. A miller sold 42 loads of flour, each load containing 9 I barrels, at 7 dollars a barrel ; how many barrels of ilour did ' he sell, and what did the whole cost ? Ans. He sold barrels ; cost, 2646 dollars. IT 23« !• A piece of valuable land, containing 33 acres, (number of rows,) was sold for 246 dollars an acre, (stars in a row ;) what did the whole cost ? ) Note 1. — When the multiplier exceeds 12, it is more convenient ' to multiply by e^h figure separately : — FIRST OPERATION. SOLUTION. — In Multiplicand^ 246 dollars, price of 1 acre, this example, the Multiplier, 33 number of acre^. J 3^ tens 'ann Ist product, 738 dollars, price of 3 acres. S^ng^i^^^'t^" 3 units, gives us 738 dollars, the price of 3 acres. Having found the price of 3 acres, our next step is to get the price of 30 acres. SECOND OPERATION. To do this, we multiply by tho 246 dollars, price of I acre. 3 tens, (thirty,) and write the 33 number of acres. first figure of the product (8) m -^ tens^ place, that is, directly under 738 dollars, price of 3 acres. the jigure by which we multiply. 738 (tens) price of SO acres, ^or the price of 30 acres being \ /- ./ ^Q times the price of 3 acres, it I 8118 dollars, price of 23 acres, will consist of the same figures, each being removed 1 place to- waids the left, by which its value is increased 10 timea. Then addr 38 MULTIH.ICA'ncm OP SIMPLE NUMBERS. IT 23 ing together the price of 3 acres, and the price of 30 acres, we hare the price of 33 acres. Note. — The correctness of the above operation results from the fact that when units (1st order) are multiplied by units, (Ist ordei,^ the product is units, Ist order. Tens (2d order) X units, (1st order,} the product is units of the 2d order. Hundreds (3d order) X tens, (2d order,) the product is units of the 4th order. 4.nd universally, — If a figxire of any order be multiplied by some figure of an- other order, the product will be units of that order indicated by the sum of their indices ^ mimes 1. Thus, 7 of the 5th order, (700G0,) multiplied by 4 of the 3d order, (400,) their indices being 5 -j- 3 = 8, and 8 — 1 = 7, their product will be 28 units of the 7th order, that is, 28 millions. 2. How many yards in 23 pieces of broadcloth, each piece containing 67 yards? OPK RATION. Solution. — Multiplying 67 yafds 67 yards in each piece, ^y 3, we get 201 yards in 3 pieces ; and 23 number o^ pieces, multiplying 67 by 2 tens, we get 134 ^ tens, = 1340 yards m 20 pieces. Add 201 yards in 3 pieces, the two products together, and we get 134 " " 20 pieces, 1^^ yards (No. of stars) in 23 pieces. — — Ans. 1640 yards. 1541 " " 2^^ pieces. Hence, — To perform multiplication when the multiplier exceeds 12, — I. Multiply the multiplicand by each figure in the multi- plier separately, first by the units, then by the tens, &c., re- membering always to place the first figure of each product directly under its multiplier. II. Having multiplied in this manner by each figure in the multiplier, add these several products together, and their swm will be the answer. Proof. — Take the multiplicand for the multiplier, and the multi- pUer for the multiplicand, and if the product be the same £is at first, the work may be supposed to be right. Qnestions, — f 23. How do you multiply when the multiplier exceeds 12? When do you write the first figure of each product? Why ? What do you do with the several products? Repeat the rule. What is the method of proof? A figure of any one order mjiltiplied by iom« tlgiirs of another ord«r, the product will b« what ? *T23. 'MULTIPLICATION OP SMPLE NUMBERS. 39 BXAMPUBS FOR PRACTICE. 3. There are 320 rods in a mile ; bow many rods are there in 57 miles ? 320 X 57 = how many? 4. It is 436 miles firom Boston to the city of Washington ; how many rods is it ? 5. What will 784 chests of tea cost, at 69 dollars a chest ? 784 X 69 = how many ? 6. If 1851 men receive 758 dollars apiece, how many dol* lars will they all receive ? Am, 1403058 dollars. Note. — Proof by the exceu of nines. Casting out the nineB in the multiplicand, we have an exoess of 6, which we write he- 1851 fore the sign of multiplication. Also, we find the excess 759 in the multiplier to be 2, which we write after the sign. The product of the nines cast out from each factor will be 1403058 an exact number of nines, since every nine multiplied by nine produces an exact number of nines. Hence, if there PROor. is an excess of nines in the entire product, it must be from 3 an excess in the pcoduct of the excesses, 6 and 2, found B X 3 in the factors. Multiplying 6 by 2, and casting out nine 3 firom the product, we write the excess, 3, over the sign ; and casting out the nines from the product of the factors, we find the excess will be the same number 3, which we write under the sign, and presume that the work is right. 7. There are 24 hours in a day ; if a ship sail 7 miles in an hour, how many miles will she sail in 1 day, at that rate ? how many miles in 36 days ? how many miles in 1 year, or 365 days ? Arts, 61320 miles in 1 year. 8. A merchant bought 13 pieces of cloth, each piece con- taining 28 yards, at 6 dollars a yard ; how many yards were there, and what was the whole cost ? Am, to the last, 2184 dollars. 9. Multiply 37864 by 235. Product,. 8898040. 10. " 29831 " 952. " 28399112. 11. « 93956 " 8704. " 817793024. 12. The factors of a certain number are 25 and 87 ; what is the number? Am, 2175. 13. A hatter sold 15 cases of hats, each containing 24 hats worth 8 dollars apiece ; how many hats did he sell, and to how many dollars did they amount ? Am, to the last, 2880 dollars. 14 A grazier sold 23 head of cattle every year for 6 years, at an average price of 17 dollars a head ; how many head of cattle did he sell, to how much did they amount each year, and to how much did they amount in 6 years ? Ans, to the last, 2346 dollars. 40 II0LTIPLICATION OF 6IMPLB TftTMBBRfl. T 24 Contractions in Multiplication. ^ 94. I. When the miUHplier is a composite number. Any number which can be produced by multiplying two or more numbers together, is called a Composite number, and The numbers which are multiplied together to produce it are called its Component parts, or Factors; thus, l5 can be produced by multiplying together 3 and 5, and is, therefore, a composite number, and the numbers 3 and 5 are its compo- nent parts. So, also, 24 is a composite number. Its component parts may be 2 and 12, (2 X 12 = 24,) or 3 and 8, (3 x 8 = 24,) or 4 and 6, (4 X 6 = 24,) or 2, 3, and 4, (2 X 3 X 4 = 24,) or 2, 2, 2, 3, (2X2X2X3 = 24.) 1. What will 18 yards of cloth cost, at 4 dollars a yard ? 3 X 6 = 18. It follows, therefore, that 3 and 6 are compo- nent parts of 18. OPERATION. Solution. — If 1 yard cost 4 dol- 4 dollars, cost of 1 yard, l^^s* ^ T^^^ will cost 3 times 4 dol- 3 yards '^^» ~ ^^ dollars ; and, if 3 yards ^ ' cost 12 dollars, 18 yards (3X6 = 12 dollars, cost of 2 yards, 18) will cost 6 times as much as 3 6 (3 X 6 =:) 18 yards. yards, that is, 6 times 12 dollars =» — 72 dollars. Hence, 72 dollars, cost of 13 yards. To perform mvltiplication when the mtdtiplier exceeds 12, and is a composite number, — I. Separate the multiplier into two or more component parts, or factors. II. Multiply the multiplicand by one of the component parts, and the product thus obtained by the other, and so on, if th? component parts be more than two, till you have multi- plied by each one of them. • The last product will be the product required. Questions. — ^T 24. What is a composite mimber? What are the componeiit parts, or factors ? Why is 15 a composite mimber ? How many factors maj a composite number have? Is 11 a composite num- ber? Why not ? Explain the 1st example. How do you multiply by a composite number? Docs it matter by which factor you multiply first ? Have you performed the 3 operations, (Ex. 2,) and compared their products 7 t26. ' MULTIPUCATIOIf OF BOfPUB NUnBICB& 4. ICXAMPLES FOR PRACTICS. 2. What will 136 tons of potash cost, at 96 dollars per ton? Let thie pupil make 3 operations, and multiply, 1st by 12 and 8 ; 2dly, by 4, 4, and 6 ; 3dly, by 96, and compare tlie operations ; he will find the results to be the same in each case. Am, 13056 dollars. 3. Supposing 342 men to be employed in a certain piece of work, for which they are to receive 112 dollars each; how much will they all receive ? 8 X 7 X 2 == 112. Am, 38304 dollars. 4. How many acres of land in 48 farms, each contaimng 367 acres ? Am. 17616 acres. '^^^ 5. Supposing 168 persons to be employed in a woollen factory, at an average price of 274 dollars each per year ; how much will they all receive ? 8x7x3 = 168. Am, 46,032 dollars. 6. Multiply 853 by 66. Product, 47,768. 7. « 18109 " 35. « 633,815. 8. " 1947271 " 81. " 157,728,951. f* 25. n. When the multiplier is 10, 100, 1000, ^-c. 1. What will 10 acres of land cost, at 25 dollars per acre ? Solution. — The price of 10 acres will be 10 times the price of 1 acre, or 10 times 25 dollars. Now if we annex a cipher to 25 dollars, price of 1 acre, oLnd the 2 tens are made 2 liun- 250 dollars, price of 10 acres, dreds. Each figure, then, is in- creased ten-fold, a^d consequently the whole number is multiplied by 10. It is also evident that ii 2 ciphers were annexed to 25, the 5 units would be made 5 hundreds, and 2 tens would be made 2 thousands, each figure being increased a hundred fold, or multiplied by 100. If 3 ciphers were annexed, each figure would be multiplied by 1000, &c Hence, When the multiplier ts 10, 100, 1000, or 1, with any num* her of cipheis annexed, — RULES. Annex as many ciphers to the multiplicand as there are Questions, — Tf 25. How are the figures of a number affected, by -l' ^rtacing one cipher at the right hand? two ciphers? three ciphers? &c. -.■li^W«tben/ do you multiply by 1, witl^ny nomber of ciphers annexed! } 492 MULTIPLICATIOir OF SBIPLE NtTMBERa ¥26 ciphers in the multiplier, and the multiplicand, so increased, will be the product required. fiXAMPLBS FOR PRACTICE. 2. What will 76 barrels of flour cost, at 10 dollars a bar- rel ? Am. 760 dollars. 3. If 100 men receive 126 dollars each, how many dollars will they al . receive ? Ans. 12600 dollars. 4. What will 1000 p.eces of broadcloth cost, estimating each piece at 312 doUi^rs ? Am. 312000 dollars. 6. Multiply 5682 by 10000. 6. « 82134 " 100000 % 96« III. When there are ciphers on the right hand of the mtdtiplicand, mtdtiplier, either or both. 1. What will 40 acres of land cost, at 27 dollars per acre ? OPERATION. Solution. — The prioe of 40 27 dollars, price of 1 acre. acres will be 40 tiroes the price ^ of 1 acre. But 40 being a com- posite number, (4X10 = 40,) 1 no J 77 ' x'A w® multiply by 4, one compo- lUW dollars, price of 4 acres. ^gnt p^rt, to get the price of 4 1080 dollars, price o/40 acres, acres, and then to multiply the price of 4 acres by 10, the other eompooenUpart, we annex a cipher to get the price of 40 acies. 2. What will 200 acres of land cost, at 400 dollars an acre ? ^ FIRST OPERATION. SOLUTION. — The 200 aCTOS 400 dollars, price of 1 acre* will cost 200 times the price 200 of 1 acre. We see in the*op- oration that the product is 8 000 with 4 ciphers at the right 000 hand, the same number as in goo the multiplicand and multi- pher counted together. We 80000 dollars, price of 200 acres, may then shorten the opera- tion, as follows : — HECOND OPERATION. Multiplying the significant figures together, 400 we place their product, 8, under the 2. Then 200 ^^ annex to this product 4 ciphers, the num- her in both factors. Hence, 80000 To perform mitltiplication when there are ciphers oh the right harid of either, or hotbUhe factor's, — \ T27. HOLTIPUCA^nON Ctf fflMPLB NBMBBBa 19 I. Set the significant figures under each oAer, placing the ciphers at the right hand. II. Multiply die significant figures together. IIL Annex as many cij^ers to the product as there are on the right hand of hoth the factors. CSXXMPLES FOR PRACTICB. 3. If 1300 men receive 460 dollars apiece, how many dol lars will they all receive ? Aru. 598000 dollars. 4. It takes 200 shingles to lay 1 course on the roof of a bam, and there are 60 courses on each of the two sides ; how many shingles will it take to cover the bam ? Am. 24000. 5. A certain storehouse contains 30 bins for storing wheat, and each bin will hold 400 bushels ; how many bushels of wheat can be stored in it ? Ans. 12000 bushels. irS7« lY. When there art dphers between the ngiuficani figures of the multiplier. 1. What is the product of 378, multiplied by 204 ? riBST OPERATION. Multiplying by a cipher second operation 378 produces nothing. There- 378 204 fore, in the muHiplication, 204 we may omit the cipher, 1512 and multiply by the sig- 1512 000 nificant figures only, as in 756 756 the second operation. Hence, to perform mal- 77112 77112 tiplication, When there are ciphers hettoeen the significant figures of the fmdtipUer, — RULE. Omit the ciphers, and multiply by the significant figures only, remembering to place the 1st figure of each product directly under its multiplier. Qne8tioif8« — Tf 26* How do you set down numbers for multipL- cation, when there are ciphers on the right hand of the multiplicand, multiplier, either or both ? How do you multiply ? How many ciphers do you annex to the product? If there were 2 ciphers on the right hand of your muUiplicand, and 5 on the right hand of your multiplier, how many would you annex to the product ? f 2T^ When there are ciphers between the significant figures of the multiplier, hpw do you multiply? Where do you set the 1st figure of the product ? 44 MULTlPUCATiOIf OF SHiPLB JNUMBEBft. T 28L EXAMPIiES FOR PRACTICE. 2. Multiply 154326 by 3007. Product, 464058282. 3. Multiply 543 by 206. Product, 111858. 4. Multiply 1620 by 2103. Product, 3406860. 5. Multiply 36243 by 32004. Product, 1159920972. 6. Multiply 101,010,101 by 1,001,001. Product, 101,111,212,111,101. IT 38. Other Methods of Contraction. I. W?ien the multiplier is 9, 99, or any number of 9% — Annex as many ciphers to the multiplicand as there are nines in the multiplier, and from the number thus produced, subtract the multiplicand ; the remainder will be the product. Thus, Multiply 6547 by 999. OPERATION. 6547000 Let the pupil prove the opeiation by actual mul* 6547 tiplication. 6540453 Ans, II. When the multiplier is 13, 14, 15, 16, 17, 18, or 19, — Multiply 32046375 by 14. Place the multiplier at the right of the mul- OPERATION. tiplicand, with the sign of multiplication be- 32046375 X 14 tween them ; multiply the multiplicand by the 128185500 *init figure of the multiplier, and set the product one place to the right of the multiplicand. This 448649250 Ans. product, added to the multiplicand, makes the true product. Note. — If the multiplier be 101, 102, <fc., to 109,— Multiply 72530486 by 103. OPERATION. 72530486 X 103 Multiply as above, and set the product two 217591458 places to the right of the multiplicand, and ~T~r — Tr~Z . ^d them together for the true product. 7470640058 A7is. ^ Questions^ — f 2S, When the multiplier is 9, 99, &;c., why does the coniraciion, as above directed, give the true product? Ans. "Multi- plying by 9 repeats the multiplicand 9 times ; annexing a cipher repeats or increases it 10 times, which is 1 time too many : hence the rule, sub- tract it 1 time, &;c. When the multiphcr is 13, 14, &c., why ? When XOl, 102, &c., why ? When 21, 31, &c., why? -^ • T29l multiplication op 8IMPLB NUMBERa 45 in. When the multiplier is 21, 31, and so on to 91, — Multiply 83107694 by 31. OPERATION. Multiply by the tens* figure only of the mul- 83107694 X 31 tiplicand, and set the unit figure of the product 249323082 under the place of the tensy and so on ; then "——--—- add them together for the true product. 2576338514 Ans. ^ f S9. Review of Multiplication. Qiie«tioiis« — What is multiplication, and how defined? Explain the use of^ the diagram of stars, and show its application to Ex. 1, 122. What must the true multiplier always be? ihe product? Why can the factors exchange places? How do you multiply by 12, or Icijs? by a number greater than 12? by a composite number? by 1 with ciphers annexed ? by any number with ciphers annexed ? when there are ciphers between the significant figures ? When units of different orders are multiplied together, of what order is the product ? EXERCISBS. 1. An army of 10700 men, having plundered a city, took 80 much money, that, when it was shared among them, each man received 46 dollars ; what was the sum of money taken ? Am. 492200 dollars. 2. Supposing the number of houses in a certain town to be 145, each house, on an average, containing two families, and each family 6 members, what would be the number of inhabitants in that town ? Ans. 1740. 3. If 46 men can do a piece of work in 60 days, how many men will it take to do it in one day ? Ans. 2760. 4. Two men depart from the same place, and travel in op- posite directions, one at the rate of 27 miles a day, the other 31 miles a day ; how far apart will they be at the end of 6 days ? ' Ans. 348 miles. 5. What number is that, the factors of which are 4, 7, 6 and 20 ? Ans. 3360. 6. If 18 men can do a piece of work in 90 days, how long will it take one man to do the same ? Ans. 1620 days. 7. What sum of money must be divided between 27 men, so that each man may receive 115 dollars ? 8. There is a certain number, the factors of which are 89 and 265 ; what is that number ? 9. What is that number, of which 9, 12, and 14 are fiic- Unrs? 10. If a carriage wheel turn round 846 times in running 46 mJhTTPUOATION OF SIMPLE NUMBEBS. T29 1 mile, how many times will it turn round in the distance from New York to Philadelphia, it being 95 miles ? Ans. 32870, 11. In one minute are 60 seconds , how many seconds in 4 minutes ? in 6 minutes ? in 20 minutes ? in 40 minutes ? Ans. to the last, 2400 seconds; 12. In one hour are 60 minutes ; how many seconds in an hour ? in two hours ? how many seconds from nine o'clock in the morning till noon ? Ans, to the last, 10800 seconds. 13. Multiply 27^^27 by 19725. Product, 5A4D6Q7 57 5. 14. Two men, A and 6, start from the same place at the same time, and travel the same way ; A travels 52 miles a day, and B 44 miles a day ; how far apart will they be at the end of 10 days ? Atis. 80 miles. 15. A farmer sold 468 pounds of pork at 6 cents a pound, and 48 pounds of cheese at 7 cents a pound, and received in payment 42 pounds of sugar at 9 cents a pound, 100 pounds of nails at 6 cents a pound, 108 yards-of sheeting at 10 cents a yard, and 12 pounds of tea at 95 cents a pound ; how many cents did he owe ? Ans, 54 cents. 16. A boy bought 10 oranges ; he kept 7 of them, and sold the others for 5 cents apiece ; how many cents did he receive ? Ans. 15 cents. 17. The component parts of a certain number are 4, 5, 7, 6, 9, 8, and 3 ; what is the number ? Ans, 181440.. 18. In 1 hogshead are 63 gallons ; hoXv many gallons in 8 hogsheads ? In 1 gallon are 4 quarts ; how many quarts in 8 hogsheads ? In 1 quart are 2 pints ; how many pints in 8 hogsheads ? Ans, to the last, 4032 pints. 19. The component parts of a multiplier are 5, 3 and 5, and the multiplicand is 118; what is the multiplier? what the product ? Ans, to the last — the product is 8850. 20. An army consists of 5 divisions, each division of 8 bri- gades, each brigade of 4 regiments, each regiment of 9 com panies, and each company of 77 men, rank and file ; the number of officers, &c., to the whole army is 42, the number belonging peculiarly to each division is 19, to each brigade 25, to each regiment 11, and to each company 14 ; how many men in the army ? Am. 133937. \ f 30. DIflSlOM OF aDIM.B NU1IB8R& 4> DIVISION OF SIMPLE NUMBERS. T 90« 1. James has 12 apples in a basket, which he distributes equally among several boys, giving them 4 apples each ; how many boys receive them ? Solution. — He can give tbe apples to as many boys as the times he can take 4 apples out of the basket, which is 3 times. Ans, 3 boys. 2. If a man travel 4 miles in an hour, in how many hours will he travel 24 miles ? Solution. — It will take him as many hours as 4 is contaioed tunes in 24. Ans, 6 hours 3. James divided 28 apples equally among 3 of his com- panioiis ; how many did he give to each ? Solution. — The 28 apples are to be divided into 3 equal parts, and oae part giveii to each boy, who will thus receive 9 apples. It will require 27 apples to give 3 boy* 9 apples each, since 9 X 3 «= 27. There will be one apple left, which must be cut into 3 equal parts, and 1 part given to each boy. Note. — If a unit, or whole thing, be divided into 2 equal parts, one of those parts is called one half; if into 3 equal parts, 1 part is called 1 third; hoo parts are called 2 thirds^ &c. If divided into 4 equal parts, one part is called 1 fourlk, or orte quarter ; 2 parts are called 2 fourths y or 2 Quarters; 3 parts, 3 fourths, or 3 quarters, Sic. Ifdivlfled into 5 equal parts, tne parts are called ^flths. If into 6 equal parts, the parts are called sixths^ &c. r 4. Seven men boiiirht a barrel of flour, each man paying an equal share ; for what part erf the barrel did 1 man pay 1 2 men ? 3 men ? 4 men ? 5 men ? 6 men ? 5. Twelve men built a steamboat, each man doing an equal share of the work ; how much of the work did 3 men do ? 5 men ? 7 men ? 9 men ? 11 men ^ 6. A boy had two apples, and gave one half an apple to each of his companions ; how many were his companions ? Ans, 4. 7. A boy divided four apples among his companions, by giving them one third of an apple each ; among how many did he divide' his apples ? Ans, 12. - Questions. — 1 30. What do you understand by 1 half of any thing or number ? 1 third ? 2 thirds ? 1 seventh ? 4 sevenths ? 6 fit teenths? 8 tenths? 5 twentieths? 9 twelfths? How many halves maka a whole one? How many thirds ? fourths? sevenths? ninths? twelfths? fifteenths? twentietlus ? dec. How many thirds make three whole ones f 48 DIJISaON OF SIMPLE mJMBElt& IT 31. 8. How many quarters in 5 oranges ? Solution. — In 1 orange there are 4 quarters, and in 5 oranges there are 5 X 4 = 20 quarters. Ans. 9. How many oranges would it take to give 12 boys one quarter of an orange each ? Arts, 3 or. 10. How much is one half of 12 apples ? Arts, 6 ap. 11. How much is one third of 12 ? 12. How much is one fourth of 12? Am. 3. 13. A man had 30 sheep, and sold one fifth of them ; how many of them did he sell ? Arts, 6. 14. A man purchased sheep for 7 dollars apiece, and paid for them all 63 dollars ; what was their number ? Arts. 9. ^31. 1. How many oranges, at 3 cents each, may be bought for 12 cents ? Solution. — As many times as 3 cents caii be taken from 12 cents, so many oranges may bo bought ; the object, therefore, is to find how many times tluree ia contained in 12. 12 cents. First orangey 3 cents. 9 We see, in this example, that 12 con- Second orange, 3 cents, ^"s 3 four times, for we subtract 3 from 12 four times, after which there is g no remainder ; consequently, subirao- rpr. ■J Q f ^^^ alone is sufficient for the operation ; i /itra orange, *5 cents. \^^^ ^^ j^^y come to the same result by "" a much shorter process, called division 3 Ans, 4 oranges. Fourth orange, 3 cents, We see from the above, that one number will be contained In another as many times as it can be subtracted from it, an^ hence, that Division is a short way of performing many subtractions of the same number. The mintiend is called the • dividend, the number which is subtracted at one time is called the divisor, and the number which indicates the number of times the sub- traction is performed is called the quotient. The cost of one orange, (3 cents,) multiplied by the number of oranges, (4,) is equal to the cost of all the oranges, (12 cents ;) 12 is, therefore, a product, and 3 one of its factors ; and to find how many times 3 is contained in 12, is to find ^ \ \ 31. DIVISION OP SIMPLE NUMBERS. 49 the other factor, which, multiplied into 3, will produce 12. Hence, the process of division consists in finding one factor of a product when the other is known. 2. A man would divide 12 cents equally among 3 children ; how many would each child receive ? Solution. — The numbers in this are the same as in the former example, but the object is difierent. In the former example, the object was to see how many times 3 oenta are contained in 13 cents ; in this, to divide 13 cents into 3 equal parts. Still the object is to find a number, which, multiplied into 3, wiQ produce 13. Thii, as in the former example, is, Ans, 4 cents. Hence Division may he defined — I. The method of finding how many times one number is contained in another of the same kind. (Ex. 1.) Or, U. The method of dividing a number into a ceptain num- ber of equal parts. (Ex. 2.) The Dividend is the number to be divided, and answers to the product in multiplication. The Divisor is the number by which we divide, and answers to one of the factors. The Quotient is the result or answer, and is the other fac- tor. When anything is left, it is called the Remainder. Note. —In the first use of division, the divisor and dividend must be of the same kind, for it would be absurd to ask how many times a number of pounds of butter is contained in a number of gallons of molasses. In the second use, the quotient is of the same kind with the dividend, for if a number of acres of land should be divided into several parts, each part will still be acres of land. Sign. — The sign of division is a short horizontal line be- tween two dots, thus -7-. This shows that the number before it is to be divided by the number after it ; thus 27 ~ 9 = 3, is read, 27 divided by 9 is equal to 3 ; or, to shorten the ex- pression, 9 in 27 3 times. Or the dividend may be written in place of the upper dot, and the divisor in place of the lower dot ; thus ^ shows that 27 is to be divided by ^ as before. Questions* — If 31, In what Way is the first example performed? How might the operation be shortened ? How often is one number con- tained in mother ? What, then, is division ? Show its relation to mul- tiplication. What is the object in the first, and what in the second, example? Define division. What is the dividend? to what does it answer in subtraction, and to what in multiplication ? What the divisor, and to what does it answer in subtraction and multiplication ? What the quotient, and to what does it answer ? Explain the divisor and divi- dend in the first use of divunon. The dividend and quotient in the lecond. 6 so DIVISION OF SIMPLE NUMBERS. Ta2. DITISION TABUL ^ Note.— The expression used by the pupil in reciting the table inay be, 3 in 3 one time, 2 in 4 two times, 4 m 12 three times, &c. 1 = 4 ■^ = 8 jyi=9 #—1 t = l t = l 1 = 1 1 = 2 f = 2 J^ = 2 ¥ — 2 1 = 3 ^ — 3 -^ 3 J^-3 ¥=4 j^— 4 a^ = 4 ¥-4 ■^ = 6 i^ = 5 ^ = 5 ¥ = 5 ■¥ — 6 V^ = 6 ^ = 6 ! iyt = 6 ¥ — 7 %l — 7 ^ = 7 1 ¥=7 ^ = 8 i^=s8 ¥ = 8 ; ¥ = 8 «f = 9 ^ = 9 ¥=9 . ¥ = 9 1 ¥ ¥ ¥ ¥ 1 2 3 4 5 6 7 8 9 ¥ ¥ ¥ ¥ ¥ ¥ ¥ ¥ 1 2 3 4 5 6 7 8 9 f-l «=1 H = i ¥ — 2 n-2 If = 2 ^ — 3 B — 3 H = 3 ¥ — 4 H — 4 « — 4 V=5 lft = 5 « — 5 ¥ — 6 *»-6 ff — 6 ¥ = 7 H = 7 H-7 ¥=8 f« — 8 f!-8 V = 9 «S-9 *f-9 if 1 2 3 4 5 6 7 8 9 :28 42 54 32 33 7,or^ 6,or^^ 9,orV^ 8,orV ll,orff: :how many? : how many? : how many? : how many ? : how many? 49 32 99 84 108 . 7, or ^ 4, or ^ 11, or ff 12, or fj: 12,orJ^ : how many ? :how many? :how many? :how many? : how many ? Note. -^The pupil should be thoroughly txerdsed in the foregoing lable. IF 33* The principles of division will be made more plain to the pupil by turning his attention to the same diagram to which it was directed while illustrating the principles of mul- tiplication, since division is the reverse of multiplication. DIAGRAM OF STARS. # # # #* # # # # # « » # In multiplication, we call the whole num* ber of stars a symbol of the product. In division, a symbol of the dividend. T 33-96. DiyiisioN of aoo^u injMB«a% 61 In mnhiplicalioix, stars in a row, and nmnber of rows, are symbols of the louitiplicand and inukiplier» which are factors of the product In diyision» stars in a row, and rows af stars, are symbols of the divisor and quotient, whidi are factors of the dividend. T3S« When the object in division is to find how many times one number, or quantity, is contained in another num- ber, or quantity^ the divisor must be of the same kind as the dividendf (stars in a row,) and the quotient will be a number telling how many times (rows of stars.) On the other hand — when the object is to divide a nninber or quantity, into a given number of equal parts, the quotiem will be of the same name or kind as the dividend, (stars in a row.) If we divide 35 apples into 5 parts, the quotient, 7 apples, will be one part, (stars in a row,) and the divisor, 5, will be a number, that is, the number of parts, (rowu of stars.) IT 34* It has been remarked that division is a short way of performing many subtractions* How often can 3 be sub- tracted frcnn 963 ? Ans. 021 times. To set down 963 and subtract 3 from it 321 times would be a long and tedious pro- cess; but by division we may decompose the number 963 thu^; 963 = 900 4- 60 -j- 3, and say 3 is contained in 9(00,) 3(00) times, in 6(0,) 2(0) times, and in 3, 1 time = 321 times, which brings us to the same result in a much shorter way. irS«S« 1. How many yards of cloth, at 3 dollars a yard, can be bought for 936 dollars ? Solution. — As many yards as 3 dollars are contained times in 936 Qaestions. — If 33. What, in the diagram of stars, may be taken as a symbol of the dividend in division ? Stars in a row and rorvs of stars are taken as symbols of what, in multiplication ? of what in division ? If the dividend be 108, the divisor 12, and the quotient 9, how would you make a diagram to correspond ? ^ 33. When the object is to find how many times one number or quantity is contained in another, the divisor will be of what name or kind? the quotient will be what? When the object is to divide a num- ber or quantity into a given number of parts, the quotient will be of what name or kind ? the divisor will be what ? ^ 34. How can you make it appear from the diagram that division IS a shorter way of performing many si rtractions ? Find on the black- board the quotient of 963 divided by 3 How would this example be performed by subtraction? G2 OITISION OF 8I1IPLB NtJMBEK& 1 35 doUan, or as many as 3 can be subtracted times from 996 ; 936 dol- lars is the diWdend, (number of stars,) 3 dollars (stars in a row) the diyisor. PREPARATION. Write the divisor on the left of the divi- Bividend, ^®i^> separate them by a curred line, and Divisor, 3 j 936 <*^^ * ^® underneath. OPERATION. We may decompose the dividend thus, 036 ^ 3 ) 936 ^^ + 30 4* 6» ^^^ divide each part separately. Beginning at the left hand, we say, 3 in 9, 3 Quotient i 312 times. This quotient 3 is 3 hundred, because the 9 which we divided is hundreds ; therefore we write it under the 9 in the place of hundreds. Proceeding to the next figure, we say, 3 in 3, 1 time, which, being 1 ten, >Ke write it in tens' place. Lastly, 3 in 6, 2 times, which, be- ing units, we write* the 2 in units' place, and the work is done. The quotient (number of rows) is 3 hundred, (300,) 1 ten, (10,) and 2 units, or 312 yards, Ans* Note. — The quotient figure will always be of the same order of units as the figure divided to obtain it. 2. 2846 -5- 2 = how many? 840-r 4=how many? 500 -2- 5 = how many ? 3. If you give 856 dollars to 4 men, how many dollars will you give to each ? OPERATION. Solution. — Write down the numbers Divisor y Dividend, as before. Divide the first figure, 8, (hun- 4 Tneriy ) 856 dollars, dreds,) in the dividend as before. Pro- ceeding to the next figure, 5, (tens,) 4 is Quotient, 214 dollars, contained 1 (ten) time in 4 of the tens, or 40, and there is 1 ten left, which, added to the 6 units, will make 16, and 4 in 16 units, 4 (units) times. Ans, 214 dollars. Here again we see that the 856 is taken in three parts, 800, 40, and 16, and each part is divided separately. When this decomposing into parts can be done in the mind, as in these examples, the process b called Short Division, It can always be done when the divisor does not exceed 12. 4. Answer the following questions after the same manner, viz., 650 -J- 5= how manv ? 8490 -?- 6 ; or, what expresses the same thing, ^^^ = how many ? 8^8*Q = how many ? ASi^lSL -= how many ? 5. What is the quotient of 14371 divided by 7 ? OPERATION. Ther£ are two other things to be learned ic 7 ) 14371 this operation. First, the divisor, 7, is net con- tained n 1, the first figrure of the dividend thea Quotient, 2053 take two figures, or so many as shall contain tbe } taa. DIYISIQN OF SIMPLB NUMBER& $S divisor, and say, 7 in 14, 3 times; we write 2 in the quotient, in thousands' place, because we divided 14 tboasands. Then, again, proceeding to the next figure, 3 in the dividend mil not contain the divisor, 7 ; to obviate this difficulty, we place a cipher in the quotient, joining the 3 to the 7 tens, calling it 37 tens, and so pro- coed. Ans, 2053. Hence, for Short Division, this general I. Write the divisor at the left hand of the dividend , sep* arate them by a line, and draw a line under the dividend, to separate it from the quotient II. Find how many times the divisor is contained in the first left hand figure or figures of the dividend, and place the result directly under the last figure of the dividend taken, for the first figure of t^e quotient. III. If there be no remainder, divide the next figure in the dividend in the same way ; but, if there be a remainder, join it to the next figure of the dividend as so many tens, and then find how many times the divisor is contained in this amount, and set down the result as before. IV. Proceed in this manner till all the figures in the divi- dend are divided. SXAMPI4ES FOR PRACTICK. 6. A man has 256 hours' work to do ; how many days will it take him, if he work 8 hours each day ? Ans, 32 days. 7. a37 0j84 7 — how many ? Ans. 215477. 8. In 1 gallon are 4 quarts ; how many gallons in 2784 quarts ? Ans. 696 gallons. 9. Seven men undertake to build a barn, for which they are to receive 602 dollars ; into how many equal parts must the money be divided ? How much will 1 part be ? 3 parts ? 5 parts ? (See IT 30.) Ans, to the last, 430 dollars. ) Qaestions. — If 35. When the dividend is large, how must it be taken ? how divided ? How is it done when the divisor does not exceed 12 ? What is the preparation ? Where do you begin the division ? If you divide units, what will the quotient be f if tens, what ? hundreds, what? If at any time you have a remainder, what do you do with it? In Ex. 5 there are two things to be learned; what is the first thing? the second thing ? What then is to be done ? How do you obviate this difficulty? What does the cipher you write in the quotient show? What is short division ? When employed ? Repeat the rule. 5* / 54 DIVISION OF SIMPLE NUMBERS. Y36 10. Divide 24108 by 12. Quotient, 2009. IT 36. 1. k man gave 86 apples to 5 boys ; how many apples did each boy receive ? Dividend, Solution. — Here, dividinff Divisor, 6)86 the number of the apples (86) — by the number of boys, (5,) we Quotient, 17 1 Remainder. iind that each boy's share would be 17 apples ; but there is \ ap- ple lefl, and this apple, which is called the remainder, is a portion of the dindend yet undivided. Wherefore this 1 apple must be divided equally among the 5 boys. But when a thing is divided into 5 equal parts, one ofthe parts is called ^, (% 30.) So each boy will have \ of an apple more, or 17^ apples in all. Arts, ll\ apples. • Note 1. — The 17 (apples) expressing whole apples, are called Integers^ that is, whole numbere. Integers are numbers expressing wliole things ; thus, 86 oranges, 4 dollars, 5 days, 75, 268, &c., are integers, or whole numbers. Note 3. — The -J (1 fifth) of an apple given to each boy, ex- pressing fart of a iUvided apple, is called a Fraction, or broken num- ber. Fractions are the parts into which a unit or whole thing mfly be divided. Thus, i (I half) of an apple, { (2 thirds) of an orange, ^ (4 sevenths) of a week, are fractions. Note 3. — A number composed of a whole number and a fraction, IS called a Mixed Number ; thus, the number 17-^ (apples) in the above example, is a mixed number, being composed of the integers 17 and the fraction -^• If we examine the fraction, we shall see, that it consists of the re- mainder (1) for its numerator, and tlie divisor (5) for its denominator. Therefore, — "> If there be a remainder, set it down at the right hand of the qtio- tient for the numerator of a fraction, under which write the divisor for its denominator. 2, Eight men drew a prize of 453 dollars in a lottery; how many dollars did each receive ? Dividend, Here, after carrying the division as far aa Divisor, 8 ) 453 possible by wholp numbers, we have a re- mainder of'^5 dollars, which, written as above Qtcotient, 56 J direcfted, gives for the answer 36 dollars and | (6 eighths) of another dollar, to each man. Qnestlons* — If 36. What are integers? fractions? a mixed num- ber? If ftiere be a remainder after division, it is a portion of what? What do fou do with it ? If you have a quotient of 23-^, what was the remfti <der ? What was t\ e divisor ? > T37. Diyision of simple number& 6A T 37. Proof. 1. 16 X 5 = 80, Prodiict, ) Multiplication and division 2. DivideTidiQO -5- 5= 16. ( are the reverse of each other. We see, in the 2d of the above examples, that the product 80 of the 1st example, divided by 5, one of its factors, brings out 16, the other factbr, and hence that division may be used to prove multiplication. We see, also, in the Ist example, that the divisor and quotient of the 2d example, multiplied togeth- er, reproduce the dividend, and henpe that multiplication may be used to prove division. Hence the To prove imdtiplieation by To prove division by frnd" division, — Divide the prod- tiplication. — Multiply the di- uct by one factor, and, if the visor and quotient toother, work be right, the quotient and if the work be ri^t, the will be the other factor. product will be equal to the dividend. NoTS 1. — To prove division^ ^ there be a remainder. Multiply the integers of the quotient by the divisor, and to the product add the re- mainder. If the work be right, their sum will be equal to the divi- dend. Example, — Divide 1145 by 7. OPEHATION. PROOF. 7 )1145 163 integers of the quotient. 163f 7 divisor. 1141 4 remainder added. ) 1145^= the dividend. Note 2. — Proof by excess of nines. Find the excess of nines in the divisor, write it before the sign of multiplication, also in the quotient, and write it after the sign ; multiply together these excesses, and write the excess of nines in their product over the sign ; subtract the remainder, if any, from the dividend, and write the excess of nines in what is left under the sign. If the numbers nnder and over the sign be alike, the work is presumed to be right, in accordance with princi- ples explained in multiplication, ^ 23, note 3. Questions. — If 3T« To what, in multiplication, does the dividend in division answer ? To what, the divisor and quotient ? How, then, ill multiplication i>roved by division? How division by multiplication! How, wnen there is a remainder 7 56 DinsioN OF smpu i numbers. 1 38 . Let the papil be required to prove t) e examples whick fi^w. BXAMPIiBS FOR PRACTICE. ^ 1. Divide 1005903360 by 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12. 2. If 2 pints make a quart, liow many quarts in 8 pints ? in 12 pints ? in 20 pints ? in 24 pints ? in 248 pints ? in 3764 pints ? in 47632 pints ? Ans, to the last, 23816 quarts. 3. Four quarts make a gallon ; how many gallons in 8 quarts ? in 12 quarts ? in 20 quarts ? in 36 quarts ? in 368 quarts ? in 4896 quarts ? in 5436144 quarts ? Am. to the last, 1359036 gallons. 4. There are 7 days in a week ; how many weeks in 365 days ? Am. 52^ weeks. 5. When flour is worth 6 dollars a barrel, how many bar- rels may be bought for 25 dollars ? how many for 50 dollars? for 487 dollars ? for 7631 dollars ? 6. Divide 640 dollars among 4 men. 640 -i- 4, or ^ = 160 dollars, Am. 7. 678 -^ 6, or AJA = how many ? ~ Am. 1 13. 8. ii^ = how many? Am. 1008. 9. i^= how many? Am. 1033f 10. ^jL — how many ? Am. 384|. 11. 2^^= how many ? 12. Ai|Ji-L= how many? 13. -aiu^^uja= how many? IT 38« 1. Divide 4478 dollars equally among 21 men. Solution. — When, as in this example, the divisor exceeds 12, ihe decomposing into parts cannot be done in the mind as in short di- vision, but the whole process must be written down at length in the following manner. OPERATION. ,,We say, 21 in 44, (hun^ds,) 2 r»- » T-i* »j /-» (hundred) times, and write 2 on the Divr. Divd. Qv^t. right hand of the dividend for the first 21 ) 4478 ( 213^ figure of the quotient. That is, we 42 1st part. have 2 hundred dollars for each of 21 ~TZ~ men, requiring 21 X 2 (hundred) = 42 ^' hundred in all. This is the first part 21 2a part. divided. The 42 hundred must now />Q be subtracted from the hundreds in the ^? rt » dividend, and we find 2 (hundred) re- DO oa part, maining, to which, bringing down the 6 Remainder L^»' **f 7)**^H ^ ^'^ ^^' J^^ ^ 27, (tens,) 1 (ten) tmie. Eadi mam > Y38. DIVISION OF SIMPLE NOHBERa 67 has DOW 10 dollars more, which require 21 tens, the second part, and taking this from the 27 tens, and bringing down the 8 units, we have 68 dollars yet to be divided. 21 in 68, 3 times, that is, each roan will have 3 dollars, which will require 63 dollars, the third part, and there are 5 dollars left. This will not give each man a whole dollar, bat ^ of a dollar. So each man has 2 hundred, 1 ten, ^j dollars ; that is, 213^^ dollars, Ans, The parts into which the dividend PROOF. jg decomposed, are 42 hundreds, which 1^^ partf 4200 dollars, contain the divisor 2 (hundred) times; 2d party 210 " 21 tens, which contain the divisor 1 2d part 63 " (ten) time ; and 63, which contaui Remainder . 5 " ^® divisor 8 (units) times, or 213 . tmies m all, and the remamder 5. 4478 << We here see that the parts added make the whole sum. This method of performing the operation is called Long Division, It consists^ writing down the whole work of di Tiding, multiplying, and subtracting. From the illustrations now given, we deduce the following RUL.E. To perform Long Division, I. Place the divisor at the left hand of the dividend, and separate them by a curved line, and draw another curved line on the right of the dividend, to separate it from the quotient. II. Take as many figures on the left of the dividend as will contain the divisor one or more times ; find how many times they contain it, and put the answer at the right hand of the dividend for the first figure in the quotient. ni. Multiply the divisor by this quotient figure, and set the product under that part of the dividend which you divided. IV. Subtract this product from the figures over it, and to the remainder bring down the next figure in the dividend. V. Divide the number this makes up as before. Continue to bring do>yn and divide until aU the figures in the dividend have been brought down and divided. Proof. — Long division maybe proved by multiplication, by the excess of nines, by adding up the parts into which the Questions* — ^ 38* What cannot be done, when the divisor ex- ceeds 12 ? Into what parts is the dividend, in the first example, decora- posed ? How, and for what, is 42 obtained ? 21 ? 63 ? Explain the proof. What is long division, and in what does it consist ? Give the rule. Name the different methods of proof. Give the substance of note 1 ; note 2 } note 3. } 68 DIVISION OF SIMPLE NtmBERS. T38 dividend is decomposed, or by subtracting the remainder from the dividend and dividing what is left by the quotient, which if the work is right, will bring the divisor. Note 1. — Having brought down a figure to the remaindCT, if the number it makes up will not contain the divisor, write a cipher in the quotient, and bring down the next figure. Note 2. — When we multiply the divisor by any quotient figure, and the product is greater than the number we divided, the quotient figure is too larger and must be diminished. Note 3. — If the remainder, at any time, be greater than the divi- sor, or eqtud to it, the quotient figure is too smalls and must be Uk creased. EXAMPLES FOR PRACTICE. 1. How many hogsheads of molasses, at 27 dollars a hogs- head, may be bought for 6318 dollars ? ^ns, 234 hogsheads. arl 2. If a man's income be 1248 dollal! a year, how much is that per week, there being 52 weeks in a year ? Ans, 24 dollars per week. 3. What will be the quotient of 153598, divided by 29 ? A71S. 5296^. 4. How many times is 63 contained in 30131 ? Ans, 478^ times ; that is, 478 times, and ^ of another time. 5. What will be the several quotients of 7652, divided by 16, 23, 34, 86, and 92 ? Am. to the last, 83^. 6. If a farm, containing 256 acres, be worth 7168 dollars, what is that per acre ? Ans. 28 dollars. 7. What will be the quotient of 974932, divided by 365? Ans. 2671-j^. 8. Divide 3228242 dollars equally among 563 men ; how many dollars must each man receive ? Ans. 5734 dollars. 9. If 57624 be divided into 216, 586, and 976 equal parts, what will be the magnitude of one of each of these equal parts ? Ans. The magnitude of one of the last of these equal parts will be 59^^^. 10. How many times does 1030603615 contain 3215? Ans. 320561 times. 11. The earth, in its annual revolution round the sun, is said to travel 596088000 miles ; what is that per hour, there being 8766 hours in a year ? Atis. 68000 miles. 12. -UJi^l^fiAa — how many ? Ans. 944581-^j^. 13. liLj^ j|o ^ = how many ? Ans. 5210||?^.* 14. AAJ^ g^y ^Q 3 1 == how many ? A71S. 108247^ij|. T39, 40. DIVISION ^ smPLB NUlfBERa 69 ¥40. Coniractioiis in Diylsion. V 30* L When the divisor is a composite number, 1. Bought 18 yards of cloth for 72 dollars ; how much wa^ thit a yard ? ' Solution. — This example is the reverse of "Eol, 1, ^ 34. It was there shown that 3 and 6 are factors of 18 (3 X 6» 18.) If the 18 yards be dl- ^^^tV^U .. ^rt • ^^^ ^"^ 3 pieces, then 3 ) 72 doUars, cost of IS yards, the cost of 1 piece would 6 ) 24 dollars, cost 1 piece = 6 yards, be one third as much as .— . the cost of 3 pieces, that Ans. 4 dollars, cost of 1 yard. »» 72-!. 3 = 24 dollars ; and the coet of 1 yard would be one sixth of the cost of 6 yards, that is, 94 -7- 6 »4 dollars, fhat is, we divide the price of 18 yards by 3, and get the price of one third of 18, or 6 jards^Uid divide the price of 6 jrards by 6,' and get vhe pric^ of 1 yard. # Ans. 4 dollars. Hence, To perform division when the divisor is a composite aumber, 1. Divide the dividend by one of the component parts, and the quotient arising from that division, by the other. 11, If the component parts be more than two. — Divide by } each of them in order, and the last quotient will be the quo- tient required. EXAAfPLBS FOR PRACTICB. 2. If a man travel 28 miles a day, how many days will it take him to travel 308 miles ? 4 X 7 = 28. Ans, 11 days. 3. J)ivide 576 bushels of wheat equally among 48 (8 X 6) men. An^, 12 bushels each. 4. Divide 1260 by 63 (=7 X 9) Quotient 20. Ans, 5. Divide 2430 by 81 (= ) " 30. Ans. 6. Divide 448 by 56 (= ) " 8. Ans. V 40« It not unfrequently happens that there are remain ders after the several divisions, as in the following example. 1. A man wished to carry 783 bushels of wheat to market , how many loads wodld he have, allowing 36 bushels to a load? Questions* — ^ 39. How may you contract the operation in di- vision when the divisor is a composite number ? Which factor should you divide by first ? Repeat the rale. 60 DiyiSIOM OF SIMPI4I NUMBERS.^ 740. Solution. — First, suppose his wheat put into barrels, each barrel containing 4 bushels. It would take as many barrels as 4 is contained times in 783. 1)783 ^ It would take 195 barrels, and leave a remainder inc Oil of 3 bushels. 195 3 rem. Next^ suppose he takes 9 barrels at each load ; 9 (barrels) X 4 (the number of bushels in each barrel) = 36 bushels at a load, and he would have as many loads as the number of times 9 barrels are con- tained in 195 barrels. 9) 195 Hence we see, that he would have 21 loads, and leave a remainder of 6 barrels ; also, a former re- 21 6 rem, mainder of 3 bushels. 4)783 bushels. The whole operation stands thus : 9)195 barrels, and 3%ushels remainder. ^ 21 loads, and 6 barrels remainder. Our object now is to find the true remainder. The last remainder, 6 barrels, multiplied by the first divisor, 4, which is the number of bushels in a barrel, gives a product of 24 bushels. To this add the first remaiuder, 3 bushels, and we have the true remainder, 27 bushels. Therefore, When there are remainders in dividing by two component parts of a number ^ to get the tiiub remainder, 1. Multiply the last remainder by the Jlrst divisor, and to the product add the first remainder j the sum will be the true remainder. II. When there are mork than two 'itisors, — Multiply each remainder, except that from the first divisor, by ail the divisors preceding the divisor which give it ; to the sum of their products add the remainder from 'ne first divisor, if any. and the amount will be the true rema»iider. 2. 5783 ^ 108 = how many ? T x 4 X 9= 108; hence, three divisors. Questions* — IT 40, When there are remainders in dividing by two coinponent parts, how do you find .he true remainder ? When there are more than two, how? if there '-e a remainUer by the fii-st divisoi only, what is the true remainder ? il by the 2d divisor and none by the 1st. how do you obtain the true rcnainder ? if by the 3d, and none by the 2d and Isi, how ? if by the lit and 3d, and none by the 2d, how » Repeat the rule. T41. DinSIOIf OF IIMPLB NUMBBRa 61 OFERATIOH 3)5783 4)1927 aW2, 1st rem. 9) 481 and 3, 2d rem. 63 and 4, 3d rem. 3d rem. 4 X 4 (2i div,) X 3 (\st div.) = 48 2d rem. 3x3 (1st div.) = 9 1st rem. 2 added 2 True rem. 59 Ans. 53^^. Note. — The remainder by the Ist diyisor, if there be no other, m the true remainder. EXAMPLES FOR PRACTICB. 3. Divide 26406 by 42 = 6 X 7 ; what will be the true remainder ? Ans, 30. 4. Divide 64^3 by 3 component parts, the continued pro- duct of which is 96 ; what will be the true remainder ? Ans, 23. 5. What is the quotient of 6811 divided by the component y parts of 81 ? Ans, 84^. 6. Divide 25431 by the component parts 3 X 4 X 8 = 96, first, in the order here given ; secondly, in a reversed order, 8, 4, 3 ; and lastly, in the order 4, 3, 8, and bring out the true quotient in each case. Quotient, 264f^. f 41 . When the divisor is 10, 100, 1000, ^c. I. A prize of 2478 dollars is drawn by 10 men ; what is each man's share ? OPERATION. Solution. — It has been shown (^[25) that an- 10 ) 2478 nexing a cipher to any number is the same as mul- ' tiplying it by 10 ; the reverse of this is equally true • nAf^ct if we cut off the right hand iigure from any number '^'TQ' it is the same as drviding it by 10 ; the figures at the left will be the quotient. The figure 8, at the right, Shorter way. being an undivi(led part, is the remainder, and may QUOT. REM. be written over tlie divisor (^ 36) thus, ^^. We 247 I 8 see that 7, which was tens before, is made units ; 4, which was hundreds, is tens, &c. On the same principle, if we cut off two figures it is the same as dividing by 100 ; if three figures, the same as mviding by 1000, &«. 6 ^ DIVISION OP SIMPLE NUMBERS. 742 Hence, To divide by 1 with any number ofcipfters annexed, RUUES. Cut off, by a line, as many figures from the right hand of the dividend as there are ciphers in the divisor. The figures at the left of the line will be the quoiient, and those at the right the remainder, BXAMPLE8 FOR PRACTICE. 2. A manufacturer bought 43604 pounds of wool in 100 days ; bow many pounds did he average each day? 42604 -f- 100 = 426t^, or 426 1 04 = 426^^ pounds, Ans. 3. In one dollar are 100 cents ; how many dollars in 42425 cents ? Ans. 424/^ ; that is, 424 dollars, 25 cents. 4. 1000 mills make one dollar ; how many dollars in 4000 mills? in 25000 mills? - — in 845000? Ans. to the last, 845 dollars. 5. In one cent are 10 mills ; how many cents in 40 mills? in 400 mills ? in 20 mills ? "- in 468 mHls ? - in 4603 mills ? ' A)is. to the last, 460-j^ cents. V 43. III. When there are ciphers on the right hand of the divisor. 1. A general divided a prize of 749346 dollars equally among an army of 8000 men; what did each receive? 8 1 000 ) 749 1 346 . Solution . — The divisor 8000 : is a composite number, of which 93 5 rem, ® *"^ l^Wi are componerl parts. Dividing what 8000 men receive c tc\j \ vv lAAA mnn by 1000, which we do by cuttirjpr 5 X^drem.) X 1000 = 5000, ^^ ^^e three right hand figures of ana 5000 + .-{46 (Ist rem.) = the dividend, we get 749 dollars 5346, true remai7ider. which 8 men will receive, with a remainder of 346 dollars ; and di- viding 749 dollars, which 8 men receive, by 8, we get what 1 man receives, which is 93 dollars, and a remainder of 5. The 5 must b6 multiplied by the first divisor, 1000, and the first remainder added to the product ; or, which is the same thing, {% 25,) the first remainder, 346, may be annexed to the 5, and we have the Ans^ 93|^i§^ dolls. Qnestions, — ^41* If we annex one cipher to any number, how does it affect it ? if two ciphers, how ? three ? 6cc. If we remove the right hand figure from any number, what is the result ? How do you divide by I with any number of ciphers annexed ? What will express the remainder ? How do you divide by 10 1 by 100 ? by 1000 ? by 10000? &c. r f 1[43. DIVISION OP SIMPLE NUMBERS. 63 Hence, When tkete are ethers on the right hand of the HvisoTf 1. Cut them off, and also, as many figures from the- right band of the dividend. n. Divide the remaining figures in the dividend by the remaining figures in the divisor. in. Annex the figures cut off frtnn the dividend to the remainder for the true remainder. KXABfPUCS FOR PRACTICB. 2. In 1 square mile are 640 square acres; how many square miles in 23040 square acres ? Ans. 36 square miles. 3. Divide 46720367 by 4200000. Quot. H^^S^Va- 4. How many acres of land can be bouffht for 346500 dol- lars, at 20 dollars per acre ? Ans. 17325 acres. 5. Divide 76428400 by 900000. Quot. 84|^MJ». 6. Divide 345006000 by 84000. Quot. 4107^^8*. 7. Divide 4680000 by 20, 200, 2000, 20000, 300, 4000, 60, 600, 70000, and 80. Ans. to 9th, 66f«ij#^. V 43. Review of Division. QnestioBS. — What is division ? In what does the process consist ? Define it. The dividend answers to what in subtraction and multiplica- tion, and why? the divisor? the quotient? In what ways is division expressed? Apply the diagram of stars to division. How does long C vision diifer from short division? Why the difference? Rule for short division — for long division. Give the different methods of proof introduced in both. To what does a remainder give rise, and how writ- ten? What are fractions? When the divisor is a composite number, how do you proceed ? How are the remainders treated ? How divide by 10, 100, Sec. ? How, wnen there are ciphers at the right hand of the divisor? EXBRCISBS. 1. An army of 1500 men, having plundered a city, took 2625000 dollars ; what was each, man's share ? Ans, 1750 dollars. 2. A certain number of men were concerned in, the pay- ment of 18950 dollars, and each man paid 25 dollars ; what was the number of men ? An>s. 758. ^Questions. — If 42. When there are ciphers on the right hand of rhe divisor, what do you do first ? How do you divide ? How do you find the true remainder ? Repeat the rule. 64 DIVISION OF SIMPI^ NUMBEB3. ¥43. S, If 7412 eggs be packed in 34 baskets, how many in a basket ? Am. 218. 4. What number must I mu^tipiy oy 135, that the product may be 505710 ? A7is. 3746. 5. Light moves with such amazing rapidity, as to pass from the sun to the earth in about 8 minutes. Admitting the distance, as usually computed, to be 95,000,000 miles, at what i^te per minute does it travel ? Ans. 11875000 miles. 6. If 2760 men can dig a certain canal in one day, how many days would it take 46 men to do the same? How many men would it take to do the work in 15 days ? in 5 days ? in 20 days ? in 40 days ? in 120 days? 7. If a carriage wheel turns round 32870 times in running from New York to Philadelphia, a distance of 95 miles, how many times does it turn in running 1 mile ? Ans, 346. 8. Sixty seconds make one minute ; how many minutes in 3600 seconds? in 86400 seconds? in 604800 seconds ? in 2419200 seconds ? 9. Sixty minutes make one hour; how many hours in 1440 minutes ? in 10080 minutes ? in 40320 min- utes ? in 525960 minutes ? 10. Twenty-four hours make a day ; how many days in 168 hours ? in 672 hours ? in 8766 hours ? 11. How many times can I subtract forty-eight from four hundred and eighty ? Atis. 10 times. 12. How many times 3478 is equal to 47854 ? Ans. 13f f f I times, 13. A bushel of grain is 32 quarts ; how many quarts must I dip out of a chest of grain to make one half (J) of a bushel * for one fourth (^) of a bushel ? for one eighth (^) of a bushel ? Ans. to the last, 4 quarts. 14. Divide 9302688 by 648. Quot. 14356. 15. Divide 1030608615 by 3215. Quot. 320561. 16. Divide 5221580 by 68705. Quot. 76. 17. Divide 2764503721 by 83000. Quot. 33307, rem. 22721. 18. If the dividend be 275868665090130, and the quotient 562916859, what was the divisor ? Ans. 490070. > ) T44. KISCELLANEOUS RXEBCISBS. W MISCELLANEOUS EXERCISES, INVOLVING THE PRINCIPLES OF THE PRECEDING RULES, V 44. The four preceding rules, viz., Addition, Subtrac- tion, Multiplication, and Division, are called the Fundamentai Rules of Arithmetic, for numbers can be neither increased nor diminished but by one of these rules ; hence, these four rules are the foundation of all arithmetical operations. EXERCISES FOR THE SLATE. L A man bought a chaise for 218 dollars, and a horse for 1*2 dollars ; what did they both cost ? 2. If a horse and chaise cost 360 dollars, and the chaise cost 218 dollars, what is the cost of the horse ? 3. If the horse cost 142 dollars, what is the cost of the chaise ? ^ 4. If the sum of 2 numbers be 487, and the greater num- ber be 348, what is the less number ? 5. If the less number be 139, what is the greater number? 6. If the minuend be 7842, and the subtrahend 3481, what* is the remainder ? 7. If the remainder be 4361, and the minuend be 7842, what is th6 subtrahend ? 8. If the subtrahend be 3431, and the remainder 4361, what is the minuend ? 9. The sum of two numbers is 48, and one of the numbers is 19 ; what is the other ? 10. The greater of two numbers is 29, and their difference 10; what is the less number? 11. The less of two numbers is 19, and their difference it 10 ; what is the greater ? 12. The sum of two numbers is 136, their difference is 28) what are the two numbers ? * { Greater number, 82. ' ( Less number, 54. IfENTAL EXERCISES. 1. When the minuend and the subtrahend are given, how do you find the remainder ? Ex. 6. Note. — The pupil may be required to give written answers to these mental exercises, or he may answer orally; in either case, let 6* 66 MISCELLANEOUS EXERCIS«:S. IT 45. him turn to the exercise for the slate to which reference is made, and let him apply it in illustration of the answer he gives. Thus — Ans. Subtract the subtrahend from the minuend, and the di&r- ence wiU be the remainder, as Ex. 6, (slate,) where the minuend and subtrahend are given to find the remainder, — we subtract the subtra- hend 3481 from the minuend 7842, and the difierence, 4361, is the remainder. 2. When the minuend and remainder are given, how do you find the subtrahend ? Ex. 7. 3. When the subtrahend and the remainder are given, how do you find the minuend ? Ex. 8. 4. When you have the sum of two numbers, and one of tkem given, how do you find the other ? Ex. 9. 6. When you have the greater of two numbers, and their difference given, how do you find the less number ? Ex. 10. 6. When you have the less of two numbers, and their dif' ference given, how do you find the greater number? Ex. 11. 7. When the sum and difference of two numbers are given, how do you find the two numbers ? Ex. 12. EXERCISES FOB THE SLATE. ir4S* 1. If the multiplicand (squares in a row) be 764, f and the multiplier (rows of squares) be 25, what will be the k product (no. of squares) ? 2. If the product (no. of squares) be 18850, and the mul- tiplicand (squares in a row) be 754, what must have been the multiplier (rows of squares) ? ^ 3. If the product (no. of squares) be 18850, and the mul- tiplier (rows of squares) be 25, what must have been the multiplicand (squares in a row).? 4. If the dividend (no. of squares) be 144, and the divi- sor (squares in a row) be 8, what is the quotient (no. of rows) ? 6. If the dividend (no. of squares) be 144, and the quo- tient (no. of rows) be 18, what must have been the divisor (squares in a row) ? 6. If the divisor (squares in a row) be 8, and the quotient (rows of squares) be 18, what must have been the dividend (no. of squares) ? 7. The product of three numbers is 5^5, and two of the numbers are 5 and 7, what is the other number ? Ans. 15. # / T4B. } mSCELLANEOUS BXERCISB8. 97 KENTAL EXEBCISES. ) When the factors are given, how do you find the product? Ex.1. When the product and one &ctor are given, how do you find the other ? Ex. 2 and 3. When the dividend and quotient are given, how do you find the divisor ? Ex. 5, When the divisor and quotient are given, how do you find the dividend ? Ex. 6. When the product of three numhers and two of them are given, how do you find the other ? Ex. 7. EXSBOISBS FOB THE SLATE. IT 9A* 1* What will he the cost of 15 pounds of hutter, It 13 cents a pound ? 2. A man bought 15 pounds of butter for 195 cents ; what nras that a pound ? 3. A man buying butter, at 15 cents a pound, paid out 195 eents ; how many pounds did he buy ? • 4. When rye is 75 cents a bushel, what will be the cost of 984 bushels ? how many dollars will it be ? 5. II 984 bushels of rye cost 738 dollars, (73800 cents,) whfd is the price oi 1 bushel ? 6. A man bought rye to the amount of 738 dollars, (73800 ^nts,) ai 75 cents a bushel ; how many bushels did he buy ? 7. If 648 pounds of tea cost 284 dollars, (28400 cents,) ^hat is the price of 1 pound ? 28400 -*- 648= how many ? MENTAL EXERCISES. 1. When the price of one pound, one bushel, &c., of any commodity is given, how do you find the cost of any number (if pounds, or bushels, &c., of that commodity? Ex. 1 and 4. L* the price of the 1 pound, &c., be in cents, in what will the ^ hole cost be ? if in dollars, what ? if in shillings ? if in pence ? &c. 2. When the cost of am/ given number of pounds, or bush- els, &c., is given, how do you find the price of one pound, or bushel, &c. ? Ex. 2, 5, and 7. In what kind of money will the answer be ? 3. When the cost of a number of pounds, &c., is given, and Bho the price of one pound, &<c., how do you find the number of pounds, &c. ? Ex. S 4Ad 6* i (j& Bns(;BLLAi^ons exsrci^bs. 747 EXERCISES FOB THE SLATE. T 4y. 1. A boy bought a number of apples ; he gave away ten of them to his companions, and afterwards bought thirty-four more, and divided one half of what he then had among four companions, who received 8 apples each ; how many apples did the boy first buy ? Let the pupil take the last number of apples, 8, and reverse the process. Ans, 40 apples. 2. There is a certain number, to which if 4 be added, and from the sum 7 be subtracted, and the difference be multiplied by 8, and the product divided by 3, the quotient will be 64; what is that number ? Arts, 27. 3. If a man save six cents a day, how many cents would he save in a year, (365 days ?) how many in 45 years ? how many dollars would it 'be ? how many cows could he buy with the money, at 12 dollars each ? A?is, to the last, 82 cows, and 1 dollar 50 cents remainder. 4. A man bought a farm for 22464 dollars ; he sold one half of it for 12480 dollars, at the rate of 20 dollars per acre ; how many acres di4 he buy ? and what did it cost him per acre ? Ans. to the last, 18 dollars. 5. How many pounds of pork, worth 6 cents a pound, can be bought for 144 cents ? 6. How many pounds of butter, at 15 cents per pound, must be paid for 25 pounds of tea, at 42 cents per pound ? 7. A man married at the age of 23 ; he lived with his wife 14 years ; she then died, leaving him a daughter 12 years of age ; 8 years after, the daughter was married to a man 6 years older than herself, who was 40 years of age when the father died ; how old was the father at his death ? Atis, 60 years. 8. The earth, in moving round the sun, travels at the rate of 68000 miles an hour ; how many miles does it. travel in one' day, (24 hours ?) how many miles in one year, (365 days ?) and how many days would it take a man to travel this last distance, at the rate of 40 miles a day ? how many years? Am. to the last, 40800 years. T4B. MISCELLANEOUS EXERCISES. 60 > • B Problems in the Measurement of Rectangles and Solids. NoTZ. — A rectangle is a figure haying four sideSy and aadi ni^ four comers a square comer. Problem I. IT 48. The length and breadth of a rectangle given^ to find the tguare contents. 1. How many square rods in a plat of ground 6 rods long and 3 rods wide ? SoLUTiOH. — A square rod is a square measuring 1 rod on each side, like one of those in the annexed dia- gram. We see from the diagram that there are as many squares in a row as there are rods on one side, and as many rows as there are rods on the other side ; that is, 5 rows of 3 squares in a row, or 3 rows of 5 squares in a row. We multiply the-number of squares in one row by the number of rows ; 5 X 3== 15 square rods, Jjis, Hence tiie Multiply the length by the breadth, and the product will be the square contents. Note. — Three times a line 5 rods long is a line 15 rods long. Hence the pupil must not fail to notice, that we multiply the number of square rods in a piece of ground 1 rod wide and of the given length by the number of rods in the width. EXAMPIiES. 2. How many square rods in a piece of ground 160 rods long (squares in a row) and 8 rods wide (rows of squares) ? Ans, 1280 square rods. 3. How many square feet in a floor 32 feet long and 23 feet wide? * Ans, 736. 4. How many yards of carpeting, 1 yard wide, will it take Questions* — ^ 48* Describe a rectangle ; a square rod. How do you determme the number of squares in a row, and the number of rows ? Give the r ile. What is the quantity really multiplied ? Wha* absurdity in Gon&idering it otherwise ? 70^ MISCELLANEOUS EX£RCII»». ir49,dQ. to cover the floors of two rooms, one 8 yards long and 7 yards wlle» and the other 6 yards long and o yards wide ? Ans. 86 yards. 6. How many square feet o( boards will it take for the floor of a room 16 feet long and 15 feet wide, if we allow 12 square feet for waste ? Ans. 252. 6: There is a room 6 yxirds long and 6 yards wide ; how many yards of carpeting, a yard wide, will be suflicient to cover the floor, if the hearth and fireplace occupy 3 square yards? Atu, 27. Problem XL IT 49* The square contents cmd width given^ to find the length, 1. What is the length of a piece of ground 3 rods wide, and containing 15 square rods ? Solution. — Id this example we have 15, the number of squares m several rows, (see the diagram, problem I.,) and 3 the number of squares in 1 row, to find the number of rows. We divide the squares in the number of rows by the squares in 1 row. Hence, Divide the square contents by the width, and the quotient will be the length. Or really, since the divisor and dividend must be of the same denomination, we divide the whole num- ber of square rods by the square rods in a piece of land 3 rods long by 1 rod wide; thus, 15 -r 3=5 rods in length, Ans. BXAMPIiES. 2. A piece of ground containing 1280 square rods, is 8 rods in width ; what is its length ? Ans, 160 rods. 3. A floor containing 736 square feet, is 23 feet wide ; what is its length ? Ans. 32 feet. Pboblem III. T ffO. The square contents and length given, to find the vndith. 1. What is the width of a piece of groundr5 rods long, and containing 15 square rods ? Questions* — ir49« Repeat the 2d problem ; the example. What two things are given in the example, and what required? Give this rule. What is really the divisor, and why ' T51. XiaCBLLAHEODS EXEECISES. 71 SoLUTioH. — We diride ihe sqture eontenia by the length, o> ' leaHy by the aquaie contents of a poition of the ground 5 rod* toag Mid 1 rod wide. Atu. 3 rods. 2. A piece of ground containing 1280 square rods, is 160 rods in length ; wltat is its width ? Am. 8 rods. 3. What is the width of a &eld ISO rods long, and con- taining 13392 squats rods ? Atu. 72 rods. Peoblbm [V. V SI. The length, breach, and hight, or thickneu given, to find, the amtenU of a solid body.* 1. How many solid feet of wood in a pile 5 feet long, 3 feet wide, and 4 feet high ? Solution. — A solid foot is a •olid 1 foot long, I wide, and 1 high. By carefully inspecting the diagram, we may see that a portion of wood 5 feet long, 1 foot wide, and 1 higb, will contain 5 solid feet. Multiplying 5 solid feet by 3, we get the contents of a portion 5 feet long, 3 feet wide, and 1 foot high. 5X3~=13 solid feet; and multi- plying 15 solid feet by 4, we get the contents of the whole pile, 16X 4 = Ml solid feet, Ans. These aia the quantities multiplied, but for le we adopt die following Huliiply the length by the breadth, and the resulting prod- uct by the bight. EXAMPLBS. 2. A laborer engaged to dig a cellar 27 feet long, 21 feet QnesIioBS. — If SO. Repeat the 3d problem ; the example ; rule. What is really the divisor, and why ! y SI. What is the 1th problem? the first eiample? Describe a solid foot. "What quantity do you maUiply in the first multiplication t inthesecond? What rule do you adopt for ■ Ths cube, or right prinn. 72 MISCELLANB0U8 EXERCISES. f 52. wide, and 6 feet deep; how many solid feet must he re- move ? A71S. 3402 solid feet. 3. A fanner has a mow of hay 28 feet long, 14 feet wide, and 8 feet high ; how man) solid feet does it contain ? Arts. 3136 solid feet Pboblem V. IT tSSt. The solid contents^ lengthy and breadth given, to find the hight~ 1. A pile of wood 5 feet long and 3 feet wide, contains 60 solid feet ; what is its hight ? Solution. — Since the divisor must be of the same denomination as the dividend, (solid feet,) we have given the solid contents of a pile 5 feet long, 3 feet wide, and several feet high, which we divide by the solid contents of a portion having the same length and breadth, and 1 foot high, to get the number of feet in the hight of the pile. Thus, 5 X 3 = 15, and 60 -f- 15 = 4 feet in hight, Ans. Hence, RUIiE* Divide the solid contents by the product of the length mul- tiplied by the breadth. EXAMPLES. 2. A man dug a cellar 27 feet long, and 21 feet wide, and removed 3402 solid feet of earth ; what was its depth ? Ans, 6 feet. 3. A mow of hay, 28 feet long and 14 feet wide, contains 3136 solid feet ; what is its hight ? Ans. 8 feet. Note. — In a similar manner we may find the breadth or the length, when the solid contents and the other two dimensions are given. 4. A pile of wood, 4 feet wide and 6 feet high, contains 360 solid feet ; what is its length ? Ans. 15 feet. 6. A stick of timber, 78 inches long and 8 inches thick, contains 6864 solid inches ; what is its width ? A?is» 11 inches. Qaestions. — If 52. What is the 5th problem ? the first example ? solution ? rule ? When the solid contents, width, and hight are given, how may the length be found? When the solid contents, length, ana hight are given, how may the width be found ? } t la-iSu llBKSLLA]«Bim8 EXOCUmB. 79 TAB* (Seneral quest99m to be antwerid mentdfyt or by Gteslate if the number of squares be 84, and the squares in a row be 14, how many^will be the rows of squares ? If the number of 'squares be 9500, and the rowt of squares be 76, how many will be the squares in a row ? Were you required to form an oblong field containing 96 square rods, what, and how many ways might you rary the figure, (rows of squares and squares in a row,) each figure to contain just 96 square rods ? There is a frame, 40 feet square and 18 feet high, the sides of which are to be covered with boards 13 feet long, 1 foot wide ; what number of these boards wiU it take, allowing only 7 feet waste ? Ans, 222 boards. A room, in a furniture warehouse, is 36 feet long and 29 feet wide ; how many tables, 3 feet square, can be set in U, leaving, a space 2 feet wide on one of the sides ? Am. 108 tables« VSi. Deflnitioiui. Integers are^ distinguished as prime, composite, even, and odd. 1. A Prime rmmher is one that cannot be divided by any number except itself and unity without a remainder ; as, 1, 2, 3, 5,% 11. NoTB. --* "fwo numbers are prime to each other, as 8 and 15, when t unit is the only number by which both of them can be divided. 2. A Composite nuviber, see IT 24. 3. An Even number is one which is exactly divisible by 2. 4. An Odd number is one which is 7u>t exactly divisible by 2. ITciS. 1. One number is a Measure of another when it divides it ivithout a remaindor* Th\is, 2 is a measure of 18 ; 5 of 45; 16 of 64. 2. A nimiber is a Common Measure of two or more num- bers when it divides each of them without a remainder. Thus 3 is a common measure of 6 and 18 ; 7 of 28 and 42 ; 4 of 1^, 20 and 32 ; 5 of 10, 15, 20, 25. 5liie8ti6ii8« — IT 54. How are integers distinguish^ t What is a prh^ number? composite number? even number ? odd number? i 74 mSGELLANEOirS BXBaSCISEa IT 56, 57 3. One number is a MvUipU of another when it can be divided by it without a remairider. Thus 8 is a multiple of 2; 15 of 5; 33 of 11. 4. A number is a Common Multiple of two or more num- bers when it can be divided by each of ^em without a re- mainder. Thus, 15 is a common multiple of 3 and 5 ; 16 of 2,4and8; 28of4and7; 54 of 2, 3, 6, 9, 18 and 27. 5. An Aliquot^ or even part, is any number which is con- tained in another number exactly 2, 3, 4, 5, &c., times. Thus, 3 is an aliquot part of 15, so also is 5. Each Of the numbers, 2, 3, 4, 6, S, and 12, is an aliquot part of 24. 6. The Reciprocal of a number is a unit^ or 1, divided by the number. Thus, | is the reciprocal of 2 ; | of 3 ; | of 4; ^ of 9, &c f 50. Oeneral Principles of Plvlslon. The value of the quotient in division evidently depends on the relative values of the dividend and divisor. Example. — Let the dividend be 24, the divisor 6, and the quotient vfeill be 4. Multiplying the dividend by 2, we in effect multiply the quotient by 2. Thus, 24 X ^ = 48, and 48 -5- 6 = 8, which is 2 times 4, the quotient of 24 -r 6. Again, dividing the divisor by 2, we in effect multiply the quotient by 2. Thus, 6 -5- 2 == 3, and 24 ^ 3 = 8, which is 2 times 4, the quotient of 24 -5- 6, the same as before. Hence, - Principle I. Multiplying the dividend, or dividing the divisor, by any number, is in effect multiplying the quotient by that number. V S7« Example as above, namely, dividend 24, divisor 6, and quotient 4 Dividing the dividend by 2, we in effect divide the quotient by 2. Thus, 24 -r- 2 = 12, and 12 ^ 6 = 2, which is equal to 1 half of ihe quotient of 24 -r- 6. Again, multiplying the divisor Dy 2, we in effect divide the quotient by 2. Thus, 6x2 = 12, and 24 -H 12 = 2, which is equal to 1 half of the quotient of 24 -r 6, the same as be- fore. Hence, J ' " ' I.I II. I I , Qnestions* — IT'S*- What is a measure? common measure ? mul- tiple? common multiple? an aliquot part? Uie reciprocal of a quantity f If 56. On what does the value of the quotient m division depenaf What is the 1st prin«pk7 I I T^8-4M). HHKSLLAnBOUS K J CM C i aML 7f PumoiPLS n. IMTidiiig the dividend, or multiplTiiig the dvnaat by any nmnber, is in efiect dividing the qaotient by that number. TSS. Example, the same as before. Multiplying both dividend and divisor by 2 does not alter the quotient. Thus, 24 X 2=48; 6 X 2 = 12; and 48 -T- 12 = 4, which is equal to the quotient of 24 -S- 6. Again, dividing both dividend and divisor by 2 does not alter the quotient Thus, 24 -s- 2=: 12; 6-^2=3; and 12-^3=4, which is equal to the quotient of 24-1-6, the same as before. Hence, PftnvciPLE m. Multift3^g or dividing both dividend and divisor by the same number does not alter the quotient. ir«B9« Example. It is required to multiply 24 by 6, and divide the product by 6. 24x6 = 144, and the product ]44-S-6=:24, which is equal to the number mumplied. Hence, Principle IV. If a number be multiplied, and the produd divided by the same number, the quotient will be the number. This result depends upon the principle that if the product be divided by the multiplier, the quotient will be the multi- plicand. — ) VOO. Cancelation. 1. How many oranges, at 4 cents apiece, can be bought for 4 dimes, or 4 ten cent pieces ? SoLU'AoN. — We multiply 10 by 4 to get the number of cents, 10 X 4 S3 40 ; then as many times as 4 is contained in 40 so many oranges can be bought. But multiplying 10 and dividing the pro duct by the same number does not change it, (^ 59 ;) hence, we ma} omit both operations, taking 10 for the result, as follows : opsRi^Tioir. Writing 10 and the multiplier 4 above, and the 10 V ^ * divisor 4 below a horizontal line, we strike out 4 i lS-C s=s 10 sbove and below the line, and we have 10 for the 4 result. Arts. 10 oranges. Note. — This process of omitting 4 is called cancelation. When we cancel a numb^, we usually draw an oblique line across it. Questions. — IT 57* What is the 2d princi^de 7 If 58. What is the 3d principle ? S 69« What is th« 4th prinsiple I « 76 msom;umaom waoBBKmoM T60l 2. A farmer sold Ifi cows for S4 doUars api^ee^ «idi todk his pay in sheep at 6 doUats i^ioee'; kew maHy sheep did he receive ? SoiiUTiON. — We see that 24 is to be multiplied by the composke number 15 = 3 X 5, and the product divided by 6. Usmg the com- ponent parts of the multipUer, we multiply 24 by 3. Now the product of 24 X 3 is to be multiplied and the lesidt iHrided by 6, wlueh c^r- ations we may omit, as follows : Writiag the nnmbeis as alveady described, we OPERATION. g^^^ out 5 bdow, and I W= 3X5 above the Iine» 3 and above 15 set the factor 3, by whidi we multii> 2^X i$ MO ply ^« Since there is no number by which to di- g — '^ vide this product, it is the result required. ^ Am. 79 sheep. 3. Multiply 165 by 33, and divide the product by 31 ; multiply the quotient by 16 and divide the product by 99 ; multiply the quotient by 62 and divide the product by &5 ; multiply the quotient by 3 and divide the product by 20. OPERATION. By cloedy in- 4 2 speoting these niim- 0iX^9X$$Xfi0 5 5 the line are canceled ft 6 except 4, 2 and 3, which must be mul- tiplied together ; and that all the fectors below the line are canceled except 5, by which the product of the remaining factors above the line is to be divided. Note 1. — It is plain that 16 above and 20 below the Hne have the factor 4 common, for 16 sp 4 X 4 and 20 3934 X 5 ; we therefore can- cel the factor 4 from 16 and 20 ; this we do if we erase the two num- bers, and write 4 the other factor of 16 over it, and 5 the other factor of 20 under it. We see also that 3, the reserved factor of 165, can eels 3, the reserved factor of 99. Note 2. — 1£ the pi^>il will perform the operations at length, of multiplying and dividing, in this examj^e, he will see how mudi is saved by cancelation. Cancelation, then, is the method of erasing, or rejecting, a factor or factors, from any number or numbers, it may be applied for shortening the operation where botli multiplication and division are required, by rejecting equal factors from the numbers to be multiplied and the divisors. 1 fop. mmmjAHBom sacBRCiaBa it I. Write down the numbers to be multiplied together akntef and the divisors beloWf a horizontal line. n. Cancel all the factors common to the numbers to be multiplied and the divisors. III. Proceed with the remaining numbers as required by the question. Note. — One factor on one side of the line will cancel only one Uk$ factor on the other si4^ IfiXAMPUBS FOR PRACTICB. 4. A man sold 35 barrels of flour at 5 dollars per bairel, and took his pay in salt at 3 dollars per barrel ; he sold the salt at 4 dollars per barrel, and took his pay in broadcloth at 7 dol- lars per yard ; he sold the broadcloth at 8 dollars per yard, and took his pay in sheep at 2 dollars a head ; he sold the sheep at 3 dollars a head, and took his pay in land at 15 dol- lars per acre ; how mapy acres of land did he purchase ? If like factors be canceled from the numbers to be multi- plied and the divisors, there will remain of the numbers to be multiplied 5 X 4 X 4=80, and of the divisors 3 ; and ^ = 26$. Am. 26 1 acres. 5. What is the quotient of36x8x4x8X2 divided by 6X5X3X4X2? Note. — The remaining factors of the numbers to be multiplied are 2, 8 and 8, and of the divisors, 5. 6. In a certain operation the numbers to be multiplied are 27, 14, 40, 8 and 6, and the divisors are 7, 10, 12 and 15 ; what is the quotient ? 9X2X2X8=»288, and288-^5=57f, Arts. 7. What is the quotient of 4 X 7 X 18 X 10 X 8 X 9, divided by 24 X 72 X 3 ? Note. — All the divisors cancel. An$, 70. 8. If the numbers to be multiplied are 14, 5, 3 and 28, and the divisors 15 and 9 ; what is the quotient ? Note. — The remaining factor of the divisors is 9. Ans, 43^* Questions. — ^ 60. If a number be multiplied And the product divided by the same number, what is the result? When such (^rations are to be performed, how may they be contracted? What is this pro- cess called? How do you indicate that a number is canceled? What IS cancelation? When may it be applied? Bepeat tne mle. Explain the operation in Ex. 5 ; in £x, 6, ^ce. 7* 78 MISCETJLANEOUS EXSRCISBS. IT 61, 62. V01* To find a comtMm divucrr of tvx) <yr more n^^ 1. Find a common divisor of 6, 9 and 12. i 6 = 3X3 The factor 3, which is common to the OPERATION. < 9 = 3X3 Beveral numbers, must be a coomion divi- ( 12 ss 3 X 4 sor of them. Hence the Separate each number into two factors, one of which shall be common to all the nuAibers. ^ The common factor will be their common divisor. EXAMPLJBS FOR PRACTICK. 2. Find a common divisor of 4, 16, 24, 36 and 8. Ans, 4. 3. Find a common divisor, or common measure, (which terms mean the same thing,) of 22, 44, 66, and 88. Ans. 11. 4. Required the length of a rod which will be a common measure of two pieces of cloth, one of them 25 feet, the other 30 feet long. Ans* 5 feet. IT 03* To find the greatest common divisor of two or more numbers. The greatest common divisor of several numbers is the greatest factor common to them, and may be found by a sort of trial. Let it be required to find the greatest common divi- sor of 128 and 160. The greatest common divisor cannot exceed the less number y for it must measure it. We will try, therefore, if the less number, 128, which measures itself, will also divide or measure 160. 128 \ 160 ( 1 ^^ ^^ ^^^' ^ ivoie, and 32 remain ; 128, 12ft therefore, is not a divisor of 160. We will now try whether this remainder be not the QQvioft/A divisor sought; for if 32 be a divisor of 128 ^^' *^® former divisor, it must also be a divisor of 160, which consists of 128 -|j- 32. 32 in 128, 4 times, toitkout any remainder. Consequently it is contained in 160 = 128 + 32, just 5 times ; . that is, once more than in 128. And as no number greater than 32, the difference of the two numbers, is contained once more in the greater, it is the greatest common divisor. Hence, Qnestioiis, — If 61. What is a common divisor of two or mora aombers ? Repeat the rale for finding iL f 63. MISClLLAlflOOB EXERCiaUb- 79 of two To find the greatest common measure qffwo numbers^ Diride the greater number by the less, and that dirisor by the remaindeTt and so on, always dividii^r the last divisor by the last remainder, till nothing remain. The last dtoisar will be the greatest common divisor required. Non 1. — When we would find the greatest oooimon divisor ot more than two nnmbrak we may first find the greatest oommon divi- sor of two numbers, uR then of that oommon divisor and one of the other nombers, and so on to the last number. Then will the greatest conmion divirar last found be the answer. Note 3. — Two numbers which are prims to each other, of ooorse, can have no oom^pon divisor greater than 1*. BXAMPIJCS FOR PBiACTICB. « 1. Apply the foregoing rule to find the greatest common divisor of 21 and 35. 2. Find the greatest common divisor of 96 and 544. Ans. 32. 3. Find the greatest common divisor of 468 and 1184. Ans. 4 4. What is the greatest common divisor of 32, &0, and 25fr? Ans. 16. V 5. What is the greatest common divisor of 75, 200, 625, ' and 150? Ans. 25. 6. A certain tract of land containing 100 acres, is 160 rods long and 100 wide ; what is the length of the longest chain that will exactly measure both its length and breadth ? M Ans. 20 rods. 7. A has 2640 dollars, B 1680 dollars, and C 756 dollars, which they agree to lay out for land at the fi;reatest price per acre that will allow each to expend the whole of his money ; what was the price per acre, and how many acres did each man buy ? Atis. a bought 220 acres, B 140 acres, and G 63 acres, at 12 dollars per acre. Questions. — Hi 62. What is the greatest common divisor of tTvo or more numbers ? Describe the process of finding it for two numbers ! rule? How foimd when the numbers are more than two ? What is tb.- greatest common measure of numbers that are prime to each ether t 80 conioN PMCtiom. f 6i. COMMON FRACTIONS. fl" •S. When whole numbers, which are called integer^, (T 36,) are subjects of calculations in arithmetic, the opera- tions-are called operations in whole numbers. But it is often necessary to make calculations in regard to parts of a thing or unit. We may not only have occasion to calculate the -price of 3 barrels, 5 barrels, or 8 barrels of ^>ur, but of one third of a barrel, two fifths of a barrel, or sevm eighths of a barrel. When a unit or whole thing is divided or broken into any number of equal parts, the parts are colhd. fractions, or broken numbers, (from the Latin ward, Jrango, I break,) If it be divided into 3 equal parts, the parts are called thirds ; if into 7 equal parts, sevenths; if into 12 «qual parts, twelfths. The fraction takes its name, or denomination, from the rtumber of parts into which the unit or whole thing is divided. If the unit or whole thing be divided into 16 equal parts, the parts are called ^zxteerUhs, and 5 of these parts would be 5 sixteenths. Fractions are of three kinds, Cotnmon, (sometimes called Vtdgar,) Decimal, and Duodecimal. Common fractions are always expressed by two numbers, one above the other, with a horizontal line between them ; thus, J, I, f The number beloip the line is called the Denominator, he- cause it gives name to the parts. The number above the line is called the JVtt^werof or, b^eause it numbers the parts. ^ The denominator shows into how many parts alhing or unit is divided ; and The numerator shows how many of these parts are con- tamed in the fraction. Thus, in the fraction f , the denomina- tor, 8, shotvs that thfe unit or whole thing is divided into 8 equal parts, and the numerator, 3, shows that 3 of these parts are contained in the fraction. The numerator, 3, numbers the parts ; the denominator, 8, gives them their denomination or Questions* — IF 63. What are integers? What fractions, att4 whence their necessity ? Whence do fractions take their name ? How many kinds of fractions ? Nanoe them. How are common fractions written ? What is the lower number called, and why ? What does it show ? What is the upper number called, and why ? What determines the size of the parts, and why? What are the terms of a fraction? What ace the terms of the fraction tW? -H? ^cc 1 Y^ 66. OOHMON IrtACTIONS. SI iMmtey ftBd shoWB ikeir ^e or magmtude ; fot if a thing be divided into 8 equal parts, the parts are but half as large as if divided into but 4 equal parts. It trill evidently take 2 eighths to make 1 fourth. The amnerator and denominator, taken together, are called the termg of the fraction. ThuS) the tenns of the fraction. ^ ar^|uid 10 ; of f , 2 and 8. . V Mf It is important to bear in mind, that fractions ariee firom division, and that the numerator may be considered a dividend, and the detionanatcr a dwisoTi and the wdue of the fraction the quotient; thus, ^ is the quotient of 1 (the nu- meiator)^ divided by 2, (the denominator ;) | is the quotient arising from 1 divided by 4 ; and | is 3 times as much, that is, 3 (Uvided by 4 ; thus, 1 fourth part of 3 is ^ same as 3 fourths of 1. Hence, a common fraction is always expressed by the rign ofdhinoih "the numerator being wntten m the place of the upper dot, and the denominator in the place of the lower dot. J expresses the quotient, of which {fSSSftS^c^SSSSi. 1. If 4 oranges be equally divided among 6 boys, what part of an orange is each boy's share ? ^ A sixth part of 1 orange is ^, and a sixth part of 4 oranges 9 ) is 4 such pieces, = f . Ans. ^ of an orange. ^ 2. If 3 apples be equally divided among 5 boys, what part ( of an apple is each boy's share ? if 4 apples, what ? if 2 ap- ples, what ? if 5 apples, what ? 3. What is the quotient of 1 divided by 3 ? of 2 by 3? of 1 by 4? of 2 by 4? of 3 by 4? of 6 by 7? -of6by8? of4by6? of 2 by 14? 4. What part of an orange is a third part of 2 oranges ? one fourth of 2 oranges? i of 3 oranges ? -J of 3 oranges? iof4? iof2? +of5? |of3? iof2? IT 6ff • A fraction being part of a whole thing, is properly less than a up it, and the numerator will be less than the de- nominator, since the denominator shows how many parts ■ -■-.,.-■ Questioiifl. -~ IF 54. From what do fractions always arise ? What may the numerator be considered? the denominator? What is the value of the fracti^? Of what is } the quotient? |? ^? ^ of Sis what pdtc df 1 ? -^ of 7 is what part of i ? By what is a common fraction wX- SB COIIMON FRACnONSL f 65. make a whale thing, and there must not be so many of the parts taken as will make a whole thing. But we call an expression written in the fractional form a fraction, though its numerator equals or exceeds the denom- inator, and its value, consequently, equals or exceeds a unit ; but since there is not a strict propriety in the name, it is called an improper fraction. Hence, m A Proper Fraction is one that is less than a unit, Wk nu- memtor being less than the denominator. An Improper Fraction is one that equals or exceeds a unit, its numerator equaling, or exceeding the denominator. Thus, •^, ^ J, are improper fractions. A Simple Fraction is a single fraction, either proper or im- proper. Thus, |, f , -J^, are simple fractions. A Compound Fraction is a fraction of a fraction, or several fractions connected by the word of. Thus, ( of {^, f of -y-, f of ^ of ^, are compound fractions. A Complex Fraction is one which has a fraction, either simple or compound, or a mixed number, for its numerator, I ^ i of 4 or for its denominator, or for both. Thus, —-, —, ' , are complex fractions. A Mixed Number, as already shown, is one composed of a whole number and a fraction. Thus, 14 J, 13}, &c., are - mixed numbers. A father bought 4 oranges, and cut each orange into 6 equal parts ; he gave to Samuel 3 pieces, to James 5 pieces, to Mary 7 pieces, and to Nancy 9 pieces ; what was each one's fraction ? Was James' fraction proper or improper ? Why ? Was Nancy's fraction proper or improper ? Why ? If an orange be cut into 5 equal parts, by what fraction is 1 part expressed ? 2 parts ? 3 parts ? 4 parts ? * 5 parts ? How many parts will make unity or a whole orange ? If a pie be cut into 8 equal pieces, and two of these pieces be given to Harry, what will be his fraction of the pie ? if 5 Questions* — IF 65. What is a proper fraction, and why so called? its value? What is an improper fmction, and why so called? When is its value a unit? When greater than a unit? Wliy? What is a simple fraction? a simple proper fraction? a simple improper fraction? a compound fraction? a complex fraction? a mixed number? Whal kind of a fraction is f of jh of </^? More quutiimnif this chanutmr. Y66. OOMMOtf FBAOTION& 83 pieces be giTen to John, what will be his finetion f what ftae> tion or part of the pie will be left? ^ V M. Reduction of Fractions. ^toduction of fractions is channng them from one form to SDomer without altering their value. To reduce an improper frac' Hon to a whole or mixed miM' her, 1. In 4 halves (|>) of an apple how many whole ap- ples? Solution. — Since 2 halves (f ) of an i^ple are equal to 1 whole apple, 4 halves ()•) are equal to as many apples as the number of times 2 halves are con- tained in 4 halves, which is 3 times. 4n5. 3 apples. 3. In f of an apple how many whole apples ? in f? inJyL? iny? in iyi? in A|a? in^? 5. How many yards in ^ of a yard? in f of a yard? inf ? in f? in J^? in JgL? in J^? in^? in^? m^^ 7. How many bushels in 8 pecks ? that is, in f of a bush- el? in J^? in V? in Jja? in a^? in-Lja? iniy^? 9. If I give 27 children \ of an orange each, how many oranges will it take ? To reduce a whole or mixed number to an improper frw> tion. 2. In 2 whole apples how many halves ? Solution. — In 2 apples ara two times as many halves as there are in 1 apple. Since there ara 2 halves ^) in 1 apple, there are 2 times 2 halves in 2 apples, ^4 halves, that is, ^, Am, 4. In 3 apples how many halves ? in 4 apples ? in 6 ap- ples? in 10 apples? in 24^ in60? in 170? m 492? 6. Reduce 2 yards to thirds, Ans, }. Reduce 2} yards to thirds. Ans, f . Re- duce 3 yards to thirds. 3\ yards. 3| yards. 5 yards. 5| yards. 6J yards. 8. Reduce 2 bushels to fourths, 2f bushels. 6 bushels. 6 J bush- els. 7| bushels. — — 25| bushels. 10. In 6} oranges how many fourths of an orange ? M OOMMON P]IA<moNB. td6. oMAATioir. It wffl take 4)27 V-;anditi. -— evident, that Ans, 6| oranges, dividing the numerator, 27, ( =s the number of parts con- tained in the fraction,) by the de- pominator, 4, ( as the number of parts in 1 orange,) will give the number of whole oranges, and the remainder, written over the de- nominator, will express the frac- tional part. Heitce, To reduce an improper fraction to a whole or mixed nuwher^ Divide the numerator by the denominator ; the quo- tient will be the whole or mixed number. opfiAAirtoM* 6f &rang€9. _4 2^ fourths in 6 oranges. 3 " cont'd in the fraction. 27 = V^, Ans. Since there are^ fourths iti 1 orange, in 6 oranges ther^ are 6 times 4 fourths = 24 fourths^ and 24 fourths -f- 3 fourths == ^7 fourths. Hence, To redude a rkixed nuinher to an improper fraction^ Multiply the whole number by the denominator of the fraction; to the product add the numerator, and write the result over the denominator. Note 1. — A whole number may be reduced to the form of an im- proper fraction, by writing 1 under it for a denominator. Note 2. — A whole number may be reduced to a fVactioa having a specified denominator, by multiplying the whole number by the given denominator, and taking the product for a nu aerator^ KXAMPLrfiS FOR PRA« TIC£. 11. In -^ of a dollar, how many dollars ? 13. In ^%^ of an hour, how many hours ? 15. In 8 1 1 3 of a shining, how many shillings ? 12. In 13^ dollars, hoi;v many sixths of a dollar ? 14. What is the impropei frattion equivalent to 23fj hours ? 16. Reduce 730^ ahillingfc to an improper fraction. Qaestions* — > \ 66* What is reduction of fractions ? To what is the value of a fraction equal ? What is the rule for reducing an im- proper fraction to a whole or mixed number ? a mixed number to an miproper fraction ? How may a whole number be reduced to the form of an miproper fVacticHi 7 How Id a fraction having a specified d«!iom- inator? T97. COttOWK FlUCndNB. Bto 17« In A}f& of m day, how 18. In IM^ day«> :iow many days ? many 24ths of a day ? il«f . Aj|i -a 3761 honfB. 19. In JJ^ of a gallon, 20. In 3^j gallons, how how many gallons ? many 4ths of a gallon ? Ans, J^Li of a gallon tss 1371 quarts. 21. Reduce fj, ^^i^, fj*, 22. Reduce l^ft, 17f#, tJ^^ -^^]f^» *• wlole or mixed S^, 4.f^, and 7^ to im- numbers. proper fractions. IT 67* To reduce a fraction to its lowest or most simple terms. If ^ of an apple be divided into 2 equal parts, it becomes |. The effect on the fraction is evidently the same as if we had multiplkd both of its terms by 2. In either case, the parts are made 2 times as many as tJiey were before ; bvi they are only HALF AS LARaB ; for it will talre 2 times as msiny fourths to make a whole one as it will take halves ; and hence it is that f is the same in value or quantity as \, } is 2 parts ; and if each of these parts be again divided into 2 equal parts, that is, if both terms of the fraction be mul- tiplied by 2, it becomes 4. Now if we reverse me above operation, and divide both terms of the fraction f by 2, we obtain its equal, f ; dividing again by 2, we obtain J, which is the most simple form of the fraction, because the terms are the lecLSt possible by which the fraction can be expressed. Hence, J = |-=s=|, and the re- verse of this is evidently true, that | = | = i. It follows, therefore, by multiplying or dividing both terms of the fraction by the sarne nwrnber^ we change its terms with' out ottering its value. (IT 58.) The process of changing | into its equal J, is called reduc' ing the fraction to its lowest terms. A fraction is said to be in its lowest terms when no number greater than 1 will divide its numerator and denominator without a remainder. 1. Reduce ^^ to its lowest terms. OPERATION. We find, by iTial, that 4 ^exactly QK measure both 128 and 160, and, dividing, 128 32 4 ^^ change the fraction to its equal f^. 4y 2=: -- rrs - Atis. Again, we find that & is a divisor commou /160 40 6 ' t» both tenns^ and, dividing, we reduoe 8 86 COMMON VRACnOfiS. T67. the fraction to its equal i, which is now in its lowest tenns, for no greater number than 1 will again measure them. Note 1. — Any fraction may eridently be reduced to its lowest terms by a single division, if we use the greatest common divisor of the two terms. Thus, we may divide by 32, which we found (^ 62) to be the greatest common divisor of 128 and 160. 32) |^==^ Ans. Hence, To reduce a fraction to its lowest terms ^ Divide both tenns of the fraction by any number which will divide them both without a remainder, and the quotients thence arising in the same manner, and so on, till it appears that no number greater than 1 will again divide them. Note 2. — A number ending with a cipher is divisible by 10. If the. two right hand figures are divisible by 4, the whole number is also. A number is divisible by 2 when it ends wi^ an even number, and by 5 when it ends with 5, or 0. bxampl.es for practicb. 2. Reduce j^ to its lowest terms. Ans» •}-. 3. Reduce f g^, ^^, j-f|, and |^ to tlieir lowest terms. Ans. t, A» h f Note 3. — Let tiie following examples be wrought by both methods ; by several divisors, and also by finding the greatest ooomion divisor. 4. Reduce |^, ^^, ;ff^, and jf|$ to their lowest terms. Ans. J, J, J, and |. 6. Reduce -^^ to its lowest tenns. Ans, \. 6. Reduce ^|^ to its lowest terms. Ans. f . 7. Reduce -^^ to its lowest terms. Ans. ^J. 8. Reduce Jfff to its lowest terms. Ans. |. Note. — The reducing of a compound fraction to a simple one will be considered in the multiplication of firactions, where it properly be- longs. The reducing of fractions to a common denominator wUl be presented in connection with addition and subtraction of fractions, in which operations only it is necessary. The reducing of complex to simple fhustions will be considered after the pupil shall be made acquainted with the division of fractions, a knowledge of which ia indispensable to understanding the operation. Questions • — If 6T« Give the illustration with the half apple. Re « verse the operation. "What follows ? What is the process mentioned, and what is it called ? When is a fraction in its lowest terms ? Explain £x. 1. How can % fraction be reduced by a single division? Role. Give the i]i>te by whicn y<m. determine by what number you divid» T 08, 69. COHMON FRAOnONB. 87 Addition and Snbtraction of Fractimui. COMMON DENOBKNATOB. IT OS. 1. A boy gave to one of his companions } of an orange, to another f , to another J ; what part of an orange did he give to all? |-|-|-}-J:= how much ? Solution. — The adding together of f > f and ^ of an oraiiffe it the same as the adding of 2 oranges, 4 oranges, and 1 or^n^e, which would make 7 oranges. The 8 i called the common denominator, as it 18 conunon to the several fractions ; and we write over it the sum of the nomerators, to express the answer. Arts, i* 2. A boy had ^ of a dollar, of which he expended -^ ; what had he left ? Solution. — -^ of a dollar is one dime, or ten cent piece. The operation, then, is to subtract 3 ten cent pieces from 7 ten cent pieces, which will leave 4 ten cent pieces, or, Ans. t^. ^- i + 1 + 4 = bow much ? } — J = how much ? , 4- 2\r + /(r + A + i» + A==l^owmuch? « — A = how much ? 5. A boy, having | of an apple, gave J of it to his sister ; what part of the apple had he left ? | — J :^ how much ? IT 99» 1. A boy, having an orange, gave | of it to his sister, and J of it to his brother ; what part of the orange did he give away ? Solution. — The fractions | and | of an orange can no more be added than 3 oranges and 1 apple, which would make neither 4 oranges nor 4 apples, as they are of different kinds, {% 12,) But if 1 orange made 2 apples, the 3 oranges would make 6 apples, and the 1 apple being added we should have 7 apples. Now ^ does make just §, and consequently f make f , to which if ^ be added we shall have the Ans, f. The denominator, 4, of the fraction J, is a factor of 8, the denominator of the fraction ^. And it each term of the fraction f be multiplied by 2, the remaining factor of 8, it will be reduced to 8^*"' (f,) without altering its value. (IT 67.) Hence, if the deTiominator of one fraction be a factor of tJu denominator of am ther fraction^ and both its terms be multi' Qnestioiis. — ^ 68* Like what, is the process of adding eighths What is the 8 called, and why f What is the tenth of a dollar? L. 88 taysmatf vM^wym, tW). plied by the remaining factor , it wHl be redticed to the same denomkaaor with the latter fmctiaTii mthmU ^trifsg^iti value. (If 58.) For example : ^ 2. How many 12ths in | ? SoLtJTioN. — The factors of 12 axe 3 and 4, the latter of which is the denominator of |, and multiplying both terms of J by 3, the other factor, we have -^j a fraction of the same value as J, but having a dif^rent denominator. Ans. ^. 3. A man has -j^ of a barrel of sugar in one cask, | in another, and | in another ; how much in all ? Solution. — The denommator 6, of the second iVaetion, is a factor of 12, the denominator of the first ; and if both tenns of ^ be multi- plied by the other factor, 2, it will become -^. Also 4, the denomi- nator of the third fraction, is a factor of 12, and if both its terms be multiplied by 3, the other factor, it will be -f^. And A I "fe'f'A s=f|.= l-J^ barrels. Arw. 4. What is the amount of J, §, and f ? Solution. — As the denominators are not factors of each other, we must take some number of which each is a factor. 36 is such a num- ber. The first denominator, 4, being a factor of 36, both terms of J may be multiplied by 9, the other factor, and we shall have •^. In like manner, both terms of f being multiplied by 6, we have |^f ; and both terms of § being multiplied by 4, we have J^ ; then, -^ -j^ The process in the above examples is called reducing fraC' turns to a common dejiominatoTj and is necessary when we wish to add or subtract those of different denominators. The common denominator, it will be perceived, must contain, as <l factor, each of the other denomtTuvtors, It is not always manifest what number will contain all the denominators. There are two methods of finding such a number. FIBST METHOD. 'fl^TO. If several numbers are multiplied together, each will evidently be a factor of the product. We have, then, the following <|aestions* — H 69* How can eighths and fourths be added? When, and how, can one fraction be reduced to the denominator of another ? Explain the third example ; the fourth. What is the process called ? When is this necessary ? What must the common denomina- tor coutatn? What is wit manifest I How many methods of finding k ? T71. MmfON vtuLcmom. 89 Multiply the numerator and denominator if each fraction by^e product of the other denominators. The several new denominators will be products of the same numbers, and, therefore, will be alike ; and the numerator and denominator of each fraction being multiplied by the same number, its value is not altered. See If 58. Note. — The common denominator of two or more fractioDS is ihm common multiple of all their denominators ; ^ee % 55. 1. Reduce |, f and f- to equivalent fr^tions having a com- mon denominator. Each term of | being multiplied by 4 X 5, or 20, we have f^. " I " * 3x5, or 15, " " 1^. " t " . " 3x4, or 12, •« *» tf. The terms of each fraction are changed, while its value is not altered. 2. £educe ^, |, {^ and ^ to equivalent fractions, having a common denominator. Ans. i|^, ^J, J^g, ^§. 3. Reduce to equivalent fractions, of a common denomina- tor, and add together, ^, f , and 1. ^ns, U+U+U=U='^U^ Amount. 4 Add tc^ether j and f. Amount, 1^}. 5. What is the amount of J + J +++i • 6. What are the fractions of a common denominator equiva ient to I and J ? Am. ^ and fj, or ^j and ^f . , SECOND METHOD. VTl. While we can always find a common denominator by the above rule, it will not always give us the least com- mon denominator. In the last example, 12 as well as 24 is a common denominator of | and f . Let us see how the 12 is obtained. One number wiH contain another having several factors, when it contains all these factors. For example, let 18 be resolved into the factors 2x3x3, which, multiplied togeth- er, will produce it. It contains 6, the factors of which, 2 and 3, are the first and second factors of 18. It also contains Questions. — If 70. What is the rule in the first method ? Whence its propriety t What is a common multiple ? Explain the first example 8* ' 90 CCttlMON FfUCTiM)ll& t72. 9, the facti rs of which, 3 and 3, are the second and third fac- tors of 18. But it will not contain 8=2 X 2 X 2, for 2 is only once a factor in 18. Now 12, the factors of which are 2 X 2 X 3» will cont&in 4 = 2 X 2, since these factors are the first and second factors of 12, It will in like manner contain 6. And it is the least numher that will contain 6 and 4, for 2 must be twice a fac- tor, or it will not contain 4, and 3 must be a factor, or it will not contain 6. Hence, no one of these factors can be spared. But 24=:2 X 2 X 2 X 3, has, it is seen, 2 three times as a factor ; so one 2 can be omitted, and we have the factors of 12 as before. We have 2 as a factor once more than neces- sary, because it is a factor in both 4 and 6. Hence, when several of the deTiominators have the saTJie factor we Tieed retain it hut oTvce in the common denominator. fl" T9. The ^process of omitting the needless factors is called getting the least common denominator of several frac- tions, and is as follows : 4.6 4 and 6 are each divided by 2 ; and the divisor and remainders being taken 2.3 for the factors of the common denomina- 2 X 2 X 3 = 12. tor, we have rejected 2 once. 1. Find the least common denominator of J, |, f , |, ^. 2 . 4 . 6 . 8 • 10 Solution. — We write the denomi- : nators in a line, and divide as here seen. 1.2.3.4. 5 By the first division, 2 existing as a ' factor in each of the five numbers, is 1 I o Q ^ rejected four times, being retained once ; as the divisor is substituted for 2X2X3X2X 5= 120 the five factors 2, which we should have had by multiplying all the num- bers together. But 2 being a factor in two of the remainders, it ia rejected once more by a second division. Ans, 120. 2. Find the least common denominator of |, ^, and ^. Questions* — IfTl. Why the necessity of a second method! When will one numher contain another ? What numbers will 18 con- tain, and why ? What will it not contain ? why ? Why will 12 contain the denominators of both | and ^ ? Why is it the least number that will contain them ? Why is a factor in 24 once more than necessary f What may then be done? r tTa COMMON rRAcnoNt. n 13 8 . 12 . 34 B. 1 . 3 4.1.1 1SX3X4»96,^i9U. TWSBS> ^mtLATHMm 3 8. 13. 34 3. 3. 6 2. 1. 3 3 8. 13. 34 4. 6. 13 3. 3. 6 3. 1 . 3 1.1.1 4X3X«-84,iin.. 3X2X3X2— 24, iln# It may be seen that the product of the factors rejected by the first operation is 24, while it is 96 by the second and third. The answer by the first is consequently four times greater, and is not the least common denominator. Care must be taken to avoid this error in practice. The divisor should not be too lar^. It may always safely, though not necessanly, be the smaOest number that can divide any two or more of the denominators without a remainder* ) NBW NUMEBATOBS. V 73* 1* Reduce the fractions \, I, }, and f to equiva- lent fractions having the least common denominator. 3.3.4.6 Solution.— Tlie new denominator being foand, as above, to be 13, the denominator, 2, of the first 1.3.3.3 fraction, has been really multiplied by 6, and to "■""""""■"" preserve the eqnality of the fraction, the numerator 1.1.3.1 must 1,0 multiplied by the same number, and <| be- 3X3X2»13 comes ^. So f =^, f ==^, and f = A, and hence the firactions are •^, -f^, -^, -^. NoTK 1. — 'Die h/dot by which the numerator of any firaction is to be multiplied, may be found by dividing the common denominator by the denominator of this firaction. Hence, — Far redtidng fractions to their lowest terms. WLVJJSU Wnte down all the denominators in a line, and divide by the smallest number greater than 1 that will divide two or more of them without a, remainder. Having written the quo- Qvestions. — ^ 72* What is ihe process called ? In the first ex ample, what factors are omitted, and what substituted, by the first divi sioat What by the second? Explain the second example. m ixyuxoN FRAcnom. T78. tiei^ and nndivUted nmal^rs beiMaA, diride M befefe ; and 80 on till there wee no twd numbers which can be divided by a number greater than 1. The continued product of the quotients and divisors will be the denominator required. Then multiply each numerator by the number by which its denominator has really been mul- tiplied. Note 2. — The least ceHUBea^teneniiiiator of two or more fractioDS is the least common multiple of all their denominators. See ^ 55. 2. Reduce ^, f , and f- to fractions having the least com- mon denominator, and add ^em together. Note 3. — In writing fhictions for addition and subtraction which have a common denominator, the numerators may be written in a line, connected by the i^ropriate signs, one line extended under them all, and the denominator written under this line but once. Thus, in the last example, 3. Reduce ^ and |> to fractions of the least common denom- inator, and subtract one from the other. Arts, y\ — A = iV' difference. 4. There are 8 pieoes of cloth, one cdntainitig .7| yards, another 13f yards, and the other 15 J yiurds; how many yards in the 3 pieces ? Before adding, reduce the fractional parts to their least eoBuaaon de- nominator ; this being done, we shall have, — Adding together ail the 24ths, vis., 18-f-20 + 7|= 7^1 21, we obtain 59, that is. If = 3Ji. We writa 13f = 13|^ J > down the fraction ^ under the other fractionSy-aBd 15^ = 15fj- j reserve the 9 integers to be carried to the amount Arts S7^jjds ®^^® other integers, making in the whole 37^, 5. There was a piece of cloth containing 34f yards, from which were taken 12f yards ; how much was there left ? 34f = 34/y We cannot take 16 twenty-fourths, (^,) 12§ == 12|j| from 9 twenty-fourths, (^) ; we must, there- A oi 17 J ^^'*» borrow 1 integer, = 24 twenty-fourths Jim. -51^ yas. ^^^ j ^j^^j^^ ^^ ^^^ ^^^ ^j . ^^ ^^ ^^^ Questions* — IT 73. In getting 12 as the common denominator of the fractions in the first example, by what number has the denominator ei f been multi^ied ? By what, then, must the numerator be muHiplted ? The same questions in regard to |; in regard to f . How is this mul- ^plier found? Oive thf rule, what is the least common multiple? What is done with the sum of the fractions in the iburth exaaot^t Explain the borrowing in the fifth example. ^"74, 75. omumx wEjuBmoBm. flB subtract ^firomf|,i»dUinrew9iMtin^; aUd tikiar IS into. gen ti^m 32 inte^nt we have fit istlagem lemflBDing. Ani» 31^. f 74.^ We have, then, /or the addition and mitruetimt qffractianSf this general RiriJi» Add and subtract their numerators when they hare a cotr. mon denominator ; otherwise, they must first be reduced ta t oommon denominator. BXAMPUBS FOfR PBACltCfi. 1. W]tati8theamountof.f,4f«and 12? Am. 17ff. 2. A man bought a ticket, and sold f of it ; what part of the ticket had he left ? Ans, f . 3. Add together }, |, i, t^, \, and J^. Ammnt^^. 4» What is the difference between \i^ and 16^ f Atu, \\' 6. From IJ take |. Eemainder, 6. From 3 take J. Rem. 2 7. From 147J take 48|. Bern, 98i. a Add together 1124, 311|, and lOOOf. Am. 1424^^. 9. Add together 14, II, 4|, ^, and |. Am. 30f . 10. From | take |. Prom f take |. 11. Whett is the difiference between | and } ? } and ^? JandJ? tandf? ^ and f ?^ f and |? 12. Howmuchisl — i? 1 — i? 1 — f? I — f^ 2 — I? 2 — ^? 2i — I? 3t — T^^? 1000 — t\>I Multiplication of Fractions. IT 75. I. To multiply a fraction by a whole numier. 1. If 1 yard of cloth cost J of a dollar, what will 2 j9ixiB cjost ? J x2 = how much ? 2. If a cow consume | of a bushel of meal in 1 day, how much will she consume in 3 days ? J X 3 = how much ? 3. A boy bought 5 cakes, at f of a dollar each ; what did he give for the whole ? ^ X ^ = how much ? 4. How much is 2 times J ? 3 times J? 2timee^ft ■ I - ,- II. I II. ■ > ' I ■ Questions. — IT 74. Give the rale. How may A be redao6d to tile Jwiominator of }? f to the denominator of 1 1 (f w.) 94 COMMON FRAcnoNa IT 75. 6. Multiply f by 3. fby2. iby7. 6. A woman gives to her son f of an apple, and to her daughter twice as much ; what part of an apple does the daughter receive ? Solution. — She gives the son 3 pieces of an apple that had been cat into 8 pieces, and she may give to the daughter twice the number of the same size, that is, 6 pieces, | X ^ = f • We multiply the numerator without changing the denominator. Or, she may give the daughter 3 pieces of an apple that had been cut into half as many, that is, 4 pieces, each piece being twice as large. We divide the denominator by 2, without changing the nu* merator, showing that, as 2 small pieces make 1 large piece, the 8 small pieces will make 4 large ones. Ans, f , or }. Hence, dividing the denominator, which is the divisor, has the same effect on the value of the fraction as multiplying the numerator, which is the dividend.. (IF 56.) Hence, there are two ways to multiply a fraction by a whcle number : — I. Divide the dcTuymruitor by the whole number, (when it can be done without a remainder,) and over the quotient write the numerator. — Otherwise, II. Multiply the numerator by the whole number, and under the product write the denominator. If then the product be ^n improper fraction, it may be reduced to a whole or mixed number. SXAMPUSS FOR PRACTICE. 1. If 1 man consume -^ of a barrel of flour in a month, how much will 18 men consume in the same time ? 6 men ? 9 men ? Ans. to the last, 1 j barrels. 2. What is the product of ^^ multiplied by 40 ? -ftV X 40 = how much ? Ans. 23| . 3. Multiply ^ by 12. by 18. by 21. Dy 36. by 48. by 60. Note 1. When ^e multipUer is a composite number, we may first multiply by one component part, and that product by the other. (^ ^4.) Thus, in the last example, the multiplier, 60, is equal to 12 X 5; therefore, ^ X 12 = ii, and j^ X 5 = ^1 = 6,^, Ans. I^iiestions, — If 75, Repeat the sixth example. Why is a fhio- tion multiplied by multiplying the numerator ? Why by dividing the denominator? Give ti^e rule. How may we proceed when the multi- pHer is a composite number ? How is a mixed number multiplied ? I s 176. OOMHON FaACnORS. 4 Haldply 5f by 7. Note 3. The mixed number may be reduced to an improper fr«MV turn, and multiplied, as in the ^nrecefung examples ; but the operation wiU usually be shorter to multiply the naction and whole number Mp- araielyf and add the results together. Thus, 7 times 5 are 35 ; and 7 times } are ^=^5^^, which, added to 35, make 40^, Ans. Or, we may multiply the fraction first, and, writing down the frac- tion, lesenre the integers, to be carried to the product of the whole nuinber. 5. What will 9^ tons of hay come to, at 17 dollars per ton ? Ans. I643V dollars. 6. If a man travel 2fy miles in 1 hour, how far will he travel in 5 hours ? in 8 hours ? in 12 hours ? — in 3 days, supposing he travel 12 hours each day ? Ans, to the last, 77f miles. T 76* IL To multiply a whole number by a fraction. 1. If 36 dollars be paid for a piece of cloth, what costs f of it? SoLunmf . — If the price vf 1 piece of cloth had been given to find the price of several pieces, we should multiply the price of 1 piece by the number of pieces, and we must consequently multiply the price of 1 piece by the fraction of a piece where the price of a firaustion Lb re- quired. The price of 1 piece, 36 dollars, must be multiplied by }. One fourth of the cloth would cost i of 36, or 9 dollars, and | would cost 3 times as much, 9 X 3= 27. Ans. 27 dollars. The product is f of the multiplicand, a part denoted by the multiplying fraction. Multiplication, therefore, ^en applied to fractions, does not always imply increase, as in whole numbers ; for, when the multiplier is less than umty^ it will always require the product to be less than the nadtiplicandt to which it would be equal if the multiplier were 1. There are two operations, a division and a multiplication. But it is matter of indifference, as it respects the resulty which of these operations precedes the other, for 36 X 3 "s- 4=27, the same as 36 -S- 4 X 3=27. Hence, To mtdtiply by a fraction, we have this RUIiE. Divide the multiplicand by the denominator of the multi- Questions, — f 76. Why must 3€ be multiplied by } ? How does the product compare with the multiplicand, aad why I Oivs &• rale. 96: ooMMxm wKAxmofm. iTTV^TSr plying fraction, and multiply the quotient by th^ numerator ; oxt when there would be a remainder by division, first multi- ply by the numerator, and divide the product b^ the denomi* nator. 2. What is the product of 90 multiplied by ^ ? Am, 45. 3. Multiply: 369 by |, 4. Multiply 45 by ^. Product, ^H^ 5. Multiply 210 by |. 6. Multiply 1326 by ^. Prod. 241^1^. Note. — As either factor may be the multiplier, (^[21,) we tasy midtiply by ^ whole number, making the fraction the multiplicand. Hence, the examples in this and ^ 75, may be perftmned by the same rule. IT 77. 1. At 40 dollars for 1 acre of land, what wiH f of an acre cost ? 40 X i = how much ? In this example, the price of 1 acre, 40 dollars, is mukiplied by the fraction of an acre, f . , Ans. 32 dollars. Hence, When the pHce of unity is given, to find the cost of any quantity, less or more than unity, Multiply the price by the quantity. SXAMPUSS FOR PRACTICE. 2. If a ship be worth 1367 dollars, what is f of it worth ? Ans. 303|- dollars. 3. What cost -f} of a ton of butter, at 225 dollars per ton ? Ans. 190^ dollars. V 78« III. To rmdtiply oTie fraction by another ^ 1. At f of a dollar for one bushel of com, what will J of a bushel cost ? f X f = liow much ? Solution. — The price of one bushel, f, is to be multiplied by the fraction of a bushel, |, (% 77.) We first divide ^ by 3, to get the price of J of a bushel. This we can do by multiplying the denominator by 3, thus making the parts of a dollar only one third as large, (15ths,) while the same number, 4, is taken. ^ -r- 3 = -^ of a dollar, the price of one <^«e8tiorii8. — IT 77. Explain the first example. What two thinga are given, and what required? Eule. f 7%. COmOK FRAOTKMIB. W ^ of a dolkx. The ^nonainator 6 of the multiplicand it multiplied by 3, the denominator of the multiplier, and 4, the numerator of the multiplicand, by 2, the numerator ot the multiplier. Hence, To multiply one fractum by anotheri RITIJB. Multiply the denominators togetiier for the denominator of the product, and the numerators for tiie nttmerator of the product By this process the multiplicand is divided by the denomi- nator of the multipl3ring fraction, and the quotient multiplied by the numerator, as in T 76. CXAMPIJB8 FOR PRACTICK. 2. Multiply j^ by f Multiply ^ b3r f Product,^. 3. At ^ of a dollar a yard, what will -^ of a yard of elotii cost? 4. At 6f dollars per barrel for flour, what will ^ of a bar^ rel cost? Note. — Mixed nimibns mint be leduced to improper fractions. 6f =V; then V X A=fH=2+H dollars, Am. 5. At 1^ of a dollar per yard, what cost 7| yards ? Am. 6^ dollars. 6. At 2| dollars per yard, what cost 6§ yards ? Am, 14ff dollars. ITTO. 1. What will i of a yard of cloth cost at | of a dollar per jrard ? Solution. — We multiply the price of 1 yard, f , by |, the frac- tion of a yard, (^ 78.) Getting the price of § of a yard is getting I of f of a dollar. § of ^ is an expression called a compoond frac- tion, (^ 65.) The reducing of a compound fraction to a simple one is, then, the same as multiplying fractions together. 2. What is f of S? f of f^? J off? 3. How much is § of ^ of f ? I of ^ we hare found to be ^, and ^of % hy the above rule is -^f^, Am, Hence, Questions. — 1 78« What is the, first ap&nfionkf Ex. 1 ? Whence its prc^ety ? Second operation ? Rule. What is done by the first operation required in the rule 7 by the second operation ? 9 96 GOMMON FRACTIONS. f 80 The word of between fractions implies their continued mul- tiplication. If there are more than two fractions, we multiply together the several numerators and the several denominators. 4. How much is -^ of f of | of f ? Ans, ^V%=^t^* 5. How much is f of § of J of | ? Ans, ^. ITS©. 1. How much is ^ of ^ off off? Since the numerators are to be multiplied together, and their product to be divided by the product of the denominators. The operation may be shortened by Cancelationt (l 60.) OPERATION. Solution. — Performing this opera- tion, as described in ^ 60, we have can- 9 i ft fi 1 ^l^d all the factors of the numerators, 5vk of ? q/" -^ of ^ = J. and have the factors 2, 2, remaining of A0 0^34 the denominators. But the numerator 229 2 = 2 X 1, the numerator 5 = 5 X 1, 3 = 3X1, &c., and we have in reality the foKStOTB 1, 1, 1, and 1, left in the numerators. 1X1X1X1=1 for the new numerator, and 2X2X1X1 = 4 for the new denomina- tor. Hence, when all the factors excepting Uie I's in the numerators or denominators cancel, the new numerator or denominator, as the case nuiy be, will be 1. BXAMPUSS FOR PRACTICE IN CAHrCfiLATIOlC. 2. |of f of f of f of ^ of J of |=how much ? Ans. ^ 3. Wh^t is the continual product of ?,• J, ^ of f . and 3^ ? NoTi. — ^The integer 7 may be reduced to the form of an improper fraction, by writing a unit under it for a denominator, thus, {-. .. Ans. 2^. 4. What is the continued product of 3* f , f of J, 2^, and H off oft? ilw5. fif. 6. Reduce | of ^ of f of | of 22| to a simple fraction. Ans. 9^. 6. A horse consumed f of J of 8 tons of hay in one win ter ; how many tons did he consume ? Ans. 2^ tons. 7 Reduce J of | of f of f of f of 1 to a simple fraction. Ans. -^^ l^iiestions. — If 79. How does it appear that we have a componnd fraction in the first example ? What does the word of between fractions imply ? "What is done when there are more than two fractions ? 1 80. Why can cancelation be applied to the multipUcation of frac- tions? Explain the process. What is done with integers when occanring with fractions 7 f 81* OOlUiON FKACnOKB. M irSO. (2.) noMiscvotn vxamt^i^bb nr this unrun- PUCATlOlf OF FRACTIONS. 1. At } ddlars per yard, what cost 4 3rard8 of cloth? — — 5yards? Gyards? Syards? 20 yards? Am. to the last, 15 dollars. 2. Multiply 148 by i. byf by A- by A- Last product, 44^. 3. If 2^ tons of hay keep 1 horse through the winter, how much will it take to keep 3 horses the same time ? 7 horses ? 13 horses ? Ans. to last, 37X tons. 4. What will Sfj barrels of cider come to, at 3 dollars per barrel ? Am. 25} dollars. 5. At 14| dollars per cwt, what will be the cost of 147 cwt ? Am. 21684 dollars. 6. A owned { of a ticket ; B owned ^ of the same ; the tkket was so lucky as to draw a prize of 1000 dollars ; what was each one's share of the money ? Ans. A's share, 600^ dollars ; B's share, 400 dollars. 7. Multiply i of I by I of 4. Product, f 8. Multiply 7^ by 2tV " l^f 9. Multiply I by 2§. " 2] 10. Multiply } of 6 by f . « 11. Multiply J of 2 by J of 4. « 3. 12. Multiply continuailly together ^ of 8, | of 7, § of 9, and Jj>( 10. * Product, 20. ^ 13. Multiply 1000000 by f . Product, 55&665i. DiTfadon of Fractions. » IT 81* L To dhide a fraction by a whoU number. 1. If 2 yards of clod^pst } of a dollar, what does 1 yard cost ? how much is | dfl^d by 2 ? « 2. If a cow consume } of a bushel of meal in 3 days, how much is that per day ? | -f- 3 =s how much ? 3. If a boy divide f of an orange among 2 boys, how much will he give each one ? 4 -5- 2 = how much ? 4. A boy bov ght 5 cakes for \^ of a dollar ; what did 1 cake cost ? -fj •:- 5 = how much ? 5. If 2 bushels of apples cost f of a dollar, what is that per bushel ? 1 bushel is th^ half of 2 bushels ; the half of f is |. Am. \ dollar L - ... im OGBOtON WBACaO^L fSL. 6. If 3 J^NToes ^soDsume ^f <rf a ton of liay in & mondi, what will 1 horse conaume in the same time I Solution. — -J^ are 12 parts ; if 3 horses oonsame 12 sooh peartoi in a month, as manythnes as 8 are eontaiued k (2, so many parts 1 horse will consume. Arts. -^^ of a ton. Hence, we divide a fraction by dividiTig the numerator with- out changing the deTwrninatoTy takuag a less number of parts of the same size. 7. A woman would divide | of a pie equally between her two children ; how much does eaoh receive I Solution. — She cannot divide the 3 pieces into 2 eqoal parts and leave them all whole. But as tiie denominator 4 shows into how many parts the pie is cnt, multiplying it by 2 is equiyalent to cutting the pie into twice as many, or 8 pieces of half the size. That is, we may cut eadi piece into 2 equal pasts, and me 1 of them to each child, who will then have the same number of pieces, ^, only half as large. -Ans. f . Hence, a fraction is divided by multiplying its denominator without changing its numerator, as the parts are made small- er, while the same number is taken. Multiplying the denominator, then, which is the divisor, has the same effect on the fraction as dividing the numerator, which is the dividend, (IF 57.) Note 1. — By comparing this, and ^ 73, we shall see that where either term of a fraction is to be multipned or divided, the contrary operation may be performed on the other teim. Hence, we have two waj^ todiMe a fraction by a whole number : — I. Divide the numerator by the whole number, (if it will contain it without a remainder,) Jsng^Hyider the quotient write the denominator. -^Otherwise, i^ n. Multiply the denominator by the whole number, and over the product write the numerator. • Questions* — IF 81* How are -f^ divided by 4? What difficulty in dividing 3 pieces of pie among 2 children ? How then may | be di- vided? Why does multiply! ig the denominator divide the fraction? In what two ways is a fractioa divided ? Apply ^ 57 to the operation. What appears from comparing this with IT 75 ? Bepeat the rule. How do you divide a fraction l^ a composite number? HoW divide a mixed number ? How, when the mixed number is large ? i rsSL eeMMDei nticnoNBi 101 8. If 7 pounds of coflfee cost f^ of a doUari what is that per pound ? f^ -5- 7 =^ how much ? Ans. ^ of a dollar. 9. If ^ of an acre produce 24 bushels, what part of an acre wiU produce 1 hoshel ? ^| -{- 24 = how much ? 10. If 12 skeins of silk cost if of a dollar, what is diat a skein? |f -S- 12 = how much ? 11. Divide I by 16. Note 3. — When the divisor is a oompostte ntunber, we csn first diyide by one component part, and the quotient thenoe arising by the other, (^ 39.) Thus, in the last example, 16 = 8 X S» >i^d I -t- 8==i, and4-7-2 = A. Ans. ^. 12. Divide^ by 12. Kvide^by21. Divide ff by 24. 13. If 6 bushels of wheat cost 4{ dollars, what is it per bushel ? Note 3. — The mixed nnmbor may be rednced to an improper frac- tion, and divided as before. ^715. ff =if of a dollar, expressing the fraction in its lowest terms. 14. Divide 4^^ dollars by 9, . . Quot -A. <i(a dollar. 15. Divide 12f by 5. ; 1 \\: : 'v . ^rtr;4^ = 2f. 16. Divide 14| by 8. • • ' ^ _ Q^^^ 1 jj. 17. Divide 184i by 7.' .. : '\'::'''y,:'4ns:iSB^. NoTS 4. — When the miked number is large, it will be most conve- nient, first, to divide the whole number, and then reduce the remainder to an improper fraction ; and, after dividing, annex the quotient of the fraction to the quotient of the whole number ; thus, in the last ex- ample, dividing 184i| by 7, as in whole numbers, we obtain 26 integers with 2j = |> remainder, and, dividing this by 7, we have -j^, and 26-[-^ = ^A- ^^' . 18. Divide 2786^ by 6. iln*. 464|. 19. flow many times is 24 contained in 7646^ ? Am. 318fH- 20. How many times is 3 contamed in 462^ ? Ans. 154J-. * IT SS* F . To divide a whole number by a fraction. 1. A man would divide 9 dollars among some poor persons, giving them | of a dollar each ; how many will receive the . money?' 102 COBOION FRACTIONS. f 83. Solution. — We wish to see how mmy limes | of a dollar is con- tained in 9 dollars. Bat CFERATION. as the divisor is 4^>", (25 9 cent pieces,) we must re- 4 dace *he dividend to 4^^, as both most be of the A*^ of a dollar, 2)36^*^^ of a dollar, same denomination, (Tf 33 ;) thus, we multiply 9 12 persons, by 4, to reduce it to 4^^, since there are 4 fourths in one dollar. Then, as many times as 3 fourths are contained in 36 fourths, so many persons will receive the money. We find the number to be 12 persons, a number greater than the dividend or number of dollars. Division, then, when applied to frac- tions, does not always imply decrease. The quotient is greater than the dividend when the divisor is less than 1, to which it is just equal when the divisor is 1. Hence, To divide a whole number by a fraction, Multiply the dividend by the denominator of the dividing fraction, (thereby reducing the dividend to parts of the same magnitude as the divisor,) and divide the product by the nu- merator. l**'\ ••-,,'*• ' BXAMl^UBS* F^B. PRACTICnS. 2. l&\f . ^ftiTy-time? * ii* J: contained in 7 ? 7 -|- J = how many? *' ' '' ■■'/•.... 3. How many times can I draw J of a gallon of wine out of a cask containing 26 gallons ? 4. Divide 3 by |. 6 by |. lObyf v5. If a roan drink -^ of a quart of beer a day, how long will 3 gaUons last him? Ans. 21^ days. 6. If 2| bushels of oats sow an acre, how many acres will 22 bushels. sow ? 22 -r- 2| =s how many times ? Note. — Reduce the mixed number to an improper fraction, 2| s=-^. . Ans. 8 acres 7. How many times is ^ contained in 6 ? Ans. f of 1 time. 8. How many times is 8| contained in 53 ? Ans. 6|^ times. l^iieBtions. — If 82* How is the principle that the divisor and divi- dend mus' be of the same denomination applied to the first example ? When IS the quotient greater than the dividend, and when equal to it? Give the rule for dividmg a whole number by a fraction. 183»84. CQMllOlf FRACnOlVSw lOA 1. Ai I of a dollar per yard, how much cloth can ba bought for 12 dollars ? Solution. —-As manr times u } of a dollar is contained m (or can be subtracted firom) 13 dollais, so many yards can be bought. Ans. 18 yards. Hence, WTien the price of unity and the price cf any quan tity are giveviy tofirii the qtiontity, Divide the price of the quantity by the price of unity. KXAMPIiES FOR PRACTICB* 2. At 4f dollars a yard, how many yards of cloth may be bought for 37 dollars ? Ans. S^ yards. 3. At -fifg of aliollar tf pound, how many pounds of tea may be bought for 84 dollars ? Ans, 90^ pounds. 4. At ^ of a dollar for building 1 rod of stone waU, how many rods may be built for 87 dollars ? 87 -h f as= how many ^es ? Ans. 104f rods. T 84* III. To divide one fraction by another. 1. At I of a dollar per bushel, how many bushels of oats can be bought for ^ of a dollar ? Solution. -r- We are to divide f by f . (If 83.) To divide by a fraction we multiply the dividend by the denominator of the dividing fraction, and divide the product by the numerator. (^ 82.) iX^=^, and ^-^2=:ff=3| bushels, Ans. Hence, RUIiE* Invert the divisor, and multiply together the two upper terms for a numerator, and the two lower terms for a denomi- nator; — for thereby Ae numerator of the dividend is multi- plied by the denominator of the divisor, and thus the dividend is multiplied by this denominator, and the denominator of the dividend is multiplied by the numerator of the divisor, and thus the dividend is divided by this numerator, as in IF 82. I^nestions. — If 83* What two things are given, and what required, Ex.1? Rule. ^ 84. How do we divide by a fraction ? How multiply a fraction ! How divide a fraction? Rule for dividing one fraction by another. What thereby is done ? 104 COMMOW PRAcnoNls. IfgS, 2. Divide J by J. Q2«>^. 1. Divide i by i- Quot, 2. 3. Divide f by |. Qmo^ 3. Divide i by ^, Qtw)^. ff ^ 4. If 44 pounds of butter serve 4 family 1 week, how many weeks wiu 36|^ pounds serve them ? Note. — The mixed ntmbeis, it Will be recollected, may be redticed to improper fractions. An$, &7§7 w6eks. 5. Divide 2J by IJ. Divide lOf by 2f i^ua. 1|. QuoL 4ff« 6. How many times is iV contained in f ? - Arts. 4 times. 7. How many times is f contain^ in 4|? Arts, llf times. 8. Divide | of | by f of f ' * Quot. 4. IT 8S. 1. If } of a yard of cloth cost f of a dodlar, what is that per yard I Solution. — Had the price of several yards been given, we woultf divide it by the nmnber of yards, to find the price of 1 yard, and« in like manner, we must divide the price of the fractioln of a yard {^ of a dollar) by the fraction of a jrard, (f ,) to find the price of 1 yard. Ans, f^ of a dollar per yard. Hence, When the price of any quantity less or more than unity is given^ tbfi/m the price of unity y Divide the price by the quaintity. BXAMPLOQS FOR PRACTICE. 2. At { of a dollar for 3| bushels of apples, what does 1 bushel cost ? Ans, J of a dollar. 3. At { of a dollar for 4| bushels of oats, what does I bushel cost ? Arvs, -^ of a dollar. T 8tS« (2*) Reduction of complex to simple fractions. 1. What simple fraction is equivalent to the complex frac- tion!? * I Solution. — Since the nmneratorof a fraction is a dividend of ' ' ' III. ' . I j Qiiestioiis. — IT 85* What two things are given, and what required, Ex.1? Give the rule. f &5. eoMMon piucTi(m& 105 ik)il(A the kttotfaimilmr is te £?iflDr, ^i^ nmf divMe } by. f, by the 4} 2. Wliat simple fraction ia equid to =^ ? oPESATion. ^ Solution. — We reduce 4} to the im- 4f ^ -^ oikf propw firacdoa -^^ which we diTide by 7, ^-i-73=^, ilnf. aooording to the rale, 1[ 81. The above ilhistrationa. are sufficient to establish the fol- lowing VLtrusu Reduce any mixed number whioh may occur in the com- plex fraction to the fractional form, or any compound fraction to a simple one, afier which divide the numerator by the de- nominator, acc<Nrding to the ordinary rules for the division of fractions. ULAMPIiBS VOWL PRAOTICB. 7JL 3. Keduce the complex fraction -^ to a sknple one. 4. What is the value of ~ ? Ans. 19. 34 5. What simple fraction is equal to -^ ? Aiu» || 6. What simple fraction is equal to ^ ? Am. ^ 7JJL 7. What simple fraction is equal to -^ ? Ans. ||f 8. What simple fraction is equal to -jp ? Ans. ^=26f . 9. What simple fraction is equal to =| ? Ans. ^• 10. What simple fraction is equal to y|? Ans. ^. 11. What simple fraction is equal to \ ^ t^ AT ^ Ans.m. •^mt^mi^^^^^^mUmam^m^ Qnestioas. — 185. (2.) How is the complex fhu^tioa, Ex. 1, le itDced to a simple one? way ? Give the nik> ror reducing complex to nmple firacticlns. 1 1 ' ' •* ^ 106 COMMON FBAcnoNa 98^86. V 8ff. (3.) PROMISCUOUS RXAMPUDf Uf THB DITI^ 8ION OF FRACtlONS. 1. If 7 lb. of sugar cost -j^ of a dollar, what is it per pound? ^ -!- 7= how much? + of -^ is how muchi 2. At ^ of a dollar for f of a barrel of cider, what is that per barrel ? Am. ^ of a dollar. 3. If 4 pounds of tobacco cost { of a dollar, what does 1 pound cost ? Am. -jfj doU. 4. If ^ of a yard cost 4 dollars, what is the price per yard? Am. 4f dollars. 6. If 14f yards cost 75 dollars, what is the price per yard? Am. 5fy dollars. 6. At 31| dollars for 10^ barrels of cider,- what is that per oarrel ? Am. 3 dollars. 7. How many times is f contained in 746 ? Am. 1989|. a Divide I of i by |. Divide | by f of f Quot. f . (fiae. 3|f . 9. Divide i of ^ by | of }. « Qtiot. ^. 10. Divide 4 of 4 by nf^. Quot. 3. 11. Divide 4| by t of 4. Quot. 2^ 12. Divide f of 4 by ^. Quot. |f . 13. Divide ^ by |. Quot. 9||. f ft 86. Review of Cmaanma Fractions. Ijtvestioiis. — What tat fractions f Whence is it that the parts into which any thing or any namber may be divided, take their name? What determines the magnitude of the parts ? Why ? How does inereas- mg the- denominator anect the value of the fraction? Increasing the numerator affects it how ? How is an improper iiaction reduced to a whole or mixed number ? How is a mixed number reduced to an im- proper fraction? a whole number? How is a fraction reduced to its most simple or lowest terms? How is a common divisor found? (IT ^^O the greatest common divisor? (f 62.} Whence the necessity of reduc- ing fractions to a common denominator ? When may one fraction be reduced to the denominalor of another? What must the common denominator be ? r^ 69.) Give the first method of finding it, and the principles on wnich it is founded ; the secopd method, and the prin- ciples. What is understood by a m^uUwle ? by a common mudtiple f bj the least common multiple ? What is the process of finding it ? (^ 72.) How are fractions addeid and subtracted? How many wajrs are there to multiply a fraction by a whole number? How does it appear, that dividing the denominator mdtipUes the fractionf How is a mixed number multiiHied? What is implied in mi^tiplying b^ a fraction? Of what operations does it consist ? When the multijdier is less than a unit, what is the prodii<*t compared with the multiplicand t What two things are r \ fas. COMMON FRACnON^i V y^ 107 / n¥eii,miid what Tcqoired inf 77t What in f 13? WbatinT85f Explain the |Minciple of multiplying one fraction bjr another. Of arvid> ing one fraction by another. How do yon mnltiply a mixed number by a mixed numb^ ? How does it appear, that m multiplying both terms of the fraction by the same nnmber) tne valoe of the fraction is not altered? How many, and what are the ways of dividing a fraction by a whole number ? How does it appear that a fraction is divided by multiplying its denominator f How does dividing by a fraction differ from mudttpltfing by a fraction ? When the divisor is less than a unit, what is the quotient compared with the dividend ? How do you divide a whole number by a fraction? . KXmELCISBS. 1. What is the amount of f and f ? of ^ and I?— - of \2\, 3§, and 4| ? Am. to the last, 20f^. 2. How much is J less 4 ? A— i? -ft — A^ 14J — 4f? 6 — 4f? «* — |of|of}? Am, to the last, ^f{. 3. What fraction is that, to which if you add f the sum willbej^? iln*. H- 4. What niAber is that, from which if you take | the remainder will be ^ ? ' Am^ ij. 5. What number is that, which being divided by f the quotient will be 21 ? Am, 15|. 6. What number is that, which multiplied by | produces \ ? Am, §. ' 7. What number is that, from which if you take § of itself the remainder will be 12 ? Ans, 20. 8. What number is that, to which if you add f of f of itself the whole will be 20 ? Ans, 12. 9. What number is that, of which 9 is the J part ? Am, 13i. 10. At § of a dollar per yard, what costs | of a yard of cloth ? Am, ^1 of a dollar. 11. At 5f dollars per barrel, what costs 18J barrels of flour ? Am, 108^ dollars. 12. What costs 84 pounds of cheese, at ^ of a dollar per pound? Ans, ll-^^^ dollars. 13. What cost 46 yards of gingham, at f of a dollar per yard ? Ans, 28 J dollars. 14. What must be paid for ^oid^ yard of velvet, at 5 dol- lars per yard ? Am. 2^ dollars. 15. If i of a pound of tea cost \^oi 9l dollar, what is that per p< und ? Ans, ^ of a dollar. 16. If 7^ barrels of pork cost 73^ dollars, what is that per oarrel ? Am, 10^ doUari^, 106 ^MttMMlttMBil^ V 8^. . #7. If 4 acres of land cost 82-^ dollars, what is that pe** acre ? Ans, 20f^ dollars. 18. At 1^ of a dollar for 3j^ bushels of lime, wnat costs 1 . bushel ? Arts. A of a dollar. 19. Paid 4} dollars for coffee, at ^ of a dollar per pound , how many pounds did I buy ? An$> 29^ pounds. 20. At If dollars per bushel, how much wheat may be bought for ^% dollars? Am. 59^ bushels. 21. If 8} yards of silk make a dress, and 9 dresses be made from a piece containing 80 yards, what will be the rem« nant left ? Ans. 1^ yards. Note. — Let the pupil reverse and prove this, and the followii^ example. 22. How many vests, containing^ f of a yard each, can be made from 22 yards of vesting ? what remnant will be left ? Ans. 25 vesta. Remnant, ^ yard. » 23. What number is that, which being jjnultiplied by 15 the product will be f ? Atis, ^, 24. What is the product of l-^Jl into ^ ? 7 &fi Ans, ^f^. 25. Which of the eleven numbers, 8, 9, 11, 12, 14, 15, 16, . 8, 20, 22, 24, have all their factors the same as factors in 72? Note. — The 72 must be resolved mto the greatest number of fac- tors possible, which are 2, 2, 2, 3, 3. In like manner, each of the other numbers must be resolved. Ans. 8, 9, 12, 18, 24. 26. What is the quotient of 4^% divided by t^^i^? ^ ^ofl6 ^ Jofl9J' Ans, lO^^^y. T 87. We have seen (IF 69) that fractions having differ- ent denominators, as thirds, sevenths, elevenths, &c., can- not be added and subtracted until they are changed to eq%ud fractions, having a common denominator — a process which is t'ften somewhat tedious. To obviate this difficulty, DecimaJ fractions liave been devised, founded on the Arabic system of notation. r87. DBcnuL FRAcnona IM B* a unit or whole thing be divided into 10 equal parts, ea^ of those parts will be 1 tenth, Aus, ^ of 1 = ^ ; and if eacn tenth be divided into 10 equal parts, the 10 tenths will make 100 parts, and each part will be 1 hundredth of a whole thing, thus, ^ of ^^ y^. In like manner, if each hun- dredth be divided into ten equal parts, the parts will be thou- sandths, ^ of ^hf = ttjW> ^^^ 80 on. Such are called Decimal Fractions, from the Latin decern, meaning ten. CommoQ fractions, then, are the common ditisions of a unit or whole thing into halves, thirds, fourths, or any num- ber of parts into which we choose to divide it. Decimal fractions are the divisions of a unit or whole thing first into 10 equal parts, then each of these into 10 other equal parts, or hundredths, and each hundredth into 10 other equal parts, or thousandths, and so on. The parts of a unit, thereby, increase and decrease in a ten fold proportion in the same manner as whole'numbers. ThefoUowmg examples will show the convenienee of deci' ^mL fractioTU, 1. Add together -^ and -f^, S(a.uTioN. — Since 1 tenth makes 10 hnndredths, we may rednoe the tenths to hundredths by annexing a cipher which, in etfS&cX, multi- ^es them by 10. Thus, f^ = 30 hundredths, (V^,) and adding 16 hundredths, (i^Ar,) we have 35 hundredths, i^^-) 2. From ^ take T^ftfe. Solution. — We reduce the 36 hnn- Vy\y = 360 thousandths, dredths to thousandths by annexing a 187 thousandths, ^her to multiply it by 10. Then sub- tracting and borrowing as m whole 173 thousandths, numbers, we have left 173 thousandths, (tWV-) Qaestions* — IT 87. What are fractions? What occasions the chief difficulty in operations with common fractions? To what has this difficulty led ? What are decimal fractions ? Why are they so called ? What are the divisions and subdivisions of a unit in decimal fractions ? and what are the parts of the 1st, 2d, &c., divisions called? With what system of notation do these divisions of a unit correspond ? What is the law of increase and decrease in the Arabic system of notation ? What, then, do you say of the increase of the whole numbers, and of the parts ? How do these divisions of a unit in decimal fractions differ from the divisions of a unit in common fractions? Give examples showing th« fapehority of decimal firactiani. 10 1 10 DECIMAL FRACTIONS. % 88 . NoTS. — The pupil wiU notice, that in tkus re4ucing the fraction ^^, the j^^ makes 60 thousandths, &nd the -j^ makes 300 thou- sandths. Notation of Decimal Fractions. IT 88. 1. Let it be required to find the amount of 325-j^ -f- l^*^ + 4^^ + tUtf' ^^^ express the fractions decimally. Solution. — Since I hundred integers make 10 tens, 1 ten 10 units, 1 unit 10 tenths, 1 tenth 10 hundredths, &c., decreasing uni- formUj from left to right, we may write down the numerators of the fractions, placing tenths after units, hundredths after tenths, and so on, in this way indicating their values without expressing their denom- inators. We place a point (*) called the Decimal point, or Separatrix^ on the left of tenths to separate the fraction from units, or whole numbers. As 10 in each right hand make 1 in the next left hand column, the adding and carrying will be the samie throughout as in whole numbers. The reducing to a com- mon denominator, it will be seen, is simply filling up the vacant places with ciphers, which are omit- ted in the first operation without aflfecting the result, each figure being written in its proper place. The denominator to a decirnal fraction, although not ex pressed, is always understood, and is 1 with as many cipherw annexed as there are places at the right hand of the points Thus, *684 in the last example, is a decimal of 3 places ; coii^ sequently 1, with 3 ciphers annexed, (1000,) is its propel denominator. (Any decimal may be expressed in the form of a common fraction by writing under it its proper denominator. Thus, *684, expressed in the form of a common fraction, is m^ Qnestions. — ^ 88* Have decimal fractions numerators and de- nominalors, and both expressed? How can you T^rite the numerators so as to indicate the value of the fraction, without expressing the denom- inator? How can the denominator be known, if it is not expressed? What is the separatrix and its use ? How can a decimal be expressed in form of a common fraction'? How manv. and what advani&ges have decimal over common fractions ? reds. s. redths. iandtbs 00 _c 'P/ 'P rt M , m lu ci hund tens. units tenth hund thous hund tens, units tenth hund thous 3 2 5^5 3 2 5'5 1 6*7 8 or, 1 6*7'8 4'3 7 9 4*3 7 *0 2 5 *0 2 5 3 4 6*6 8 4 3 4 6*6 8 4 DBCniAL PRACnOlfB. Ill Note. — Turn 4edmal poiat ean never be nibly omitted in open^ tioos with decimals. Hence, Decimal fractions hare two adrantages orer common frac* tions. First J — They arc more readily reduced to a common denominator. Second, — They may be added and subtracted like whole numbers, without the formal process of reducing them to a common denominator. The names of the places to ten-miilionths, and, generally how to read or write decimal fractions, may be seen from the following ' TABUB. ^ ^ ( CD O ? i 3d 2d 1st 1st 2d 3d 4th 5th 6th 7th place, place, place. place, place, place, place, place, place, place. a>\\ CO 10 II II II oco OOO CD OOCJI ooco OlO CD II II II H«''<J^»- Tens. Units. o o oi Tenths. ca o\ Hundredths, o Thousandths. Ten-Thousandths. * Hundred-Thousandths MilHontbs. Ten-Millionths <D 8g sr C^ Ol Ol He 2 Kg g 5^5 C P'W CD . ^ 5 o s« ^ 5 S r i^tm ■ — ^-— "^j*- 112 DECIMAL mEUCTIONd. 189;;^ IT S©*. ^ read DeeiiHaU.^^At every fWustion had a nu- merator and a denominator, to read the decimal fraction, of which the denominator is not expressed, requires two en%aner* ations, — one from left to right to ascertain the denominatoi;, that is, the nanpye or denopuna^ion of the parts, and another from right to left, to ascertain the numerator, that is, the num- ber of patts. Take, for instance, the fraction *00387« We begin at the first place on the right hand of the decimal point and say, as in the table, tenths, hundredths,; &c., to the last fi^ie, which we ascertain to be hundred-thousandths, and that is the name or denomination of the fraction. Then, to know how many there are of this denomination, that is, to determine the nu- merator, we begin as in whole numbers, and say units, (7,) tens, (8,) hundreds, (3,) which being the highest significant figure, we proceed no further ; and find that we have 387 hundred-thousandths, (i (ftVoo ») ^^ numerator being 387. In this way a mixed number may be read as a fraction. Take 25^634. Beginning at the first place at the right of l^e point we have tenths, hundredths, thousandths, (the lowest denomination ;) then beginning at the right with 4, we say, units, tens, &c., as in whole numbers, and find that v^e hare 25634 thousandths, which, expressed as a common fraction, is IT 90. To write Decimal Fractions. I. Write the given decimal in such a manner that each figure contained in it may occupy the ^lace corresponding to its value. II. Fill the vacant places, if any, with ciphers, and put the decimal point in its proper place. Forty-six and seven tenths = 46^^ =46*7. Write the following numbers in the same manner : Eighteen and thirty-four hundredths. Fifty-two and six hundredths. Nineteen and four hundred eighty-seven thouBandths. Twenty and forty-two thousandths. One and five thousandths. 135 and 3784 ten-thousandths. 9000 and 342 ten-thousandths. Questions. — ^ 80. To read decimals requires what ? how made and for what purposes? How may a mixed number be read as a fnu tion? T 90 What is the mle for writing decimal fraetions f f 91. MSDntAL ItlUdnOKB. 11$ 10000 noi 15 lefr-AotMHidtlli. 974 and 108 mittifmtkB. 320 and 3 tenths, 4' hundredths, and 3 thousandths. 500 and 5 hundied-thousandths. 47 millionths. Foot hundred and twenty-three thousandths. RedQctioii of Decimal Fractioo*. IT 91. The value of every figure is determined by its place from units. Consequently, ciphers annexed to decimals do not alter their ^alue, since every significant figure continues to possess tRe same place from unity. Thus, *5, *50, *^RX), are all of the same value, each being equal to ^, or ^. But every cipher prefixed to a decimal diminishes it tenfoldy by removing the significant figures one place further from unity, and consequently making each part only one tenth as large. Thus, *5, *05, *005, are of different value, '5 being equal to ^, or | ; *05 being equal to x^, or ^ ; and *005 being equal to n^, or ^. A whole number is reduced to a decimal by annexing ci- phers ; to tenths by annexing 1 cipher, since this is multiply- ing by 10; to hundredths by annexing two ciphers, &:c. Thus, if 1 cipher be annexed to 25 it will be 25*0, (250 tenths ;) if 2 ciphers, it will be 25*00, (2500 hundredths.) Several numbers may be reduced to decimals, having the same or a common denominator, by annexing ciphers till all have the same number of decimal places. Thus, 15'7, '75, 12' 183, 9'0236 and 17* are reduced to ten thousandths, the lowest denominator contained in them, as follows : 15' 7 == 15*7000, armexing three ciphers. '75 = *7500, ** two *** QnestioDS. — IF 91. How is the value of every decimal figure de- termined? How do ciphers at the left of a decimal affect its value ? at ♦he right, how? In the fraction <02643, what is the value of the 4 ? of the 2 ? How does the affect the value of the fraction ? In the fraction '15012 is the value of each significant figure affected by the 0? if not, pwnt out the difference, and wherefore ? If from the fraction '8634 we withdraw the 6, leaving the fraction to consist of the other three figures only, how much should we deduct from its value? Demonstrate by some process that you are right. How may integers be reduced to deci- mals? Reduce 46 to thousandths; how many thousandths does the number make? How may several numbers be reduced to decimals having a common denominator? To what denominator should they all be reduced? 10* iU racniAL FRACTIONS. %9St. 12*183 = 12*1830, annexing one eii^eF. 9*0236 = 9*0236, already ten-thousandths, 17* :^ 17*0000, annexing four ciphers. NoTX. — All the numbers should be reduced to the denominatcnr of the one haying the greatest number of decimal places. BXAMPUBS. 1. Reduce 7*25, 14*082, 2*3, *00083, and 25 to a common denominator. 2. Reduce 2*1, 3*02, *425, 32*98762, and *3000001, to a common denominator. IT 93. To change or redtice Common to Decimal Fractions. 1. A man has | of a barrel of flour ; what is that, expressed in decimal parts ? As many times as the denominator of a fraction is contained in the numerator, so many whole ones are contained in the fraction. We can obtain no whole ones in f , because the denominator is not contained in the numerator. We may, however, reduce the numerator to tenths^ by annexing a cipher to it, (which, in eflect, is multiplying it by 10,) making 40 tenths, or 4*0. Then, as many times as the denominator, 5, u contained in 40, so many tenths are contained in the frac- tion. 5 into 40 goes 8 times, and no remainder. Ans, *8 of a bushel. 2. Express } of a dollar in decimal parts. OPERATION. The numerator, 3, reduced to Num. tenths, is f^, 3*0, which, divided Denom. 4 ) 3*0 ( *75 of a dMar. ^7 *.® denominator, 4, the q«o- OQ tient 18 7 tenths, and a remainder ^ of 3. This remainder must now be reduced to hundredths by an- 20 nexing another cipher, making 20 20 hundredths. Then, as many times as the denominator, 4, is contained in 20, so many hundredths also may be obtained. 4 into 20 5 times, and no remainder. | of a dollar, therefore, reduced to deci- mal parts is 7 tenths and 5 hundredths ; that is, *75 of a dollar. Questions* — ^ 02* To what is the value of every fraction equal * (IT 64.) How is a common fraction reduced to a decimal ? Of how many places must the quotient consist ? When there are not so many places how is the deficiency to be supplied ? Repeat the rule. What is the course of reasoning advanced to establish this rule ? T92. OBCniAL FRA6TIQML Hi 3. Reduce A to a dedmal fractioii. The numerator must be reduced to hundredthtf by <»»Ti#iriny two ciphers, before the diviaioii can begin. 66 ) 4«00 ( '0606 -{-, t^ il9ini»r. 396 ^ • _ As there can be no tenthi^ a AQQ cipher most be placed in the quo- qgg tient, in tentha' place. Note. — -^ cannot be reduced exactly; for, howerer long the division be oontinned , there will still be a remainder.* It is sufficiently exact for most purposes, if the decimal be extended to three or four 0aces. * Decimal figures, which eonUnually repeat, like *06, in this example, are called RepeteauUy or Circulating Decimals. If only one Jtgure repeats, as 'SSSa or *7777, &e., it is called a nngU rohetend. If two or marejmxea cir- culate altermitely, as *060606, '234234234, «c., it is called a eompoundrepetend. If ether figures arise before those which circulate, as *743333, *i48010i01, &c., the decimal is called a mixed repetend, A single repetend is denoted Dy writing only the circulating JIgvre with a pdnt over it : thus, % signifies tluit the 3 is to be continually repeated) forming an infinite or never-ending aeriea of 3s. A eompotmd repetend is denoted by a point over ihtjlret and latt repeating Jlgures: thus, *234 signifies that 234 is to be continually repeated. It may not be amiss, her6, to show how the value of any rep^tetid may be finrnd, or, in other words, how it may be reduced to itt equivalent wJigaxfraC' turn. If we attempt to reduce 4 to a decimal^ we obtain a continual repetition of the figure 1 : thus, 41111, that is, the repetend 4. The yalue of the repetend 4, then, is ^ ; the value of '222, &c., the repetend % will evidently be twice as much, that is, ^. In the same manner, 3 =^, and '4 =|^, and <6 = |^, and so on to 9, whichs -|-=: 1. 1. What is the value of *8 1 Ana. f . 2. What is the value of *6 ? Ans, f = J. What is the value of *3 7 of<7? — ^of*47 of»5? of '9? of 4? If ^ be reduced to a decimal, it produces '01 01 01 , or the repetend 'oi . The repetend '02, being 2 times as much, must be ^ and '03= ^, and '48, being 48 times as much, must be ^, and '74 =b^, &c. If ^0> be reduced to a decimal, it produces '001 ; consequently, <6o2s gfy, and '037 =^ft-» ^^^ *425 =|ff , &c. As this principle will apply to any number of places, we have this general Rulb for reducing a circulating deknuU to a common fraction. Make the given repetend the numerator, and the denominator will be a<« aiany 9s as there are repeating Jlguree, ^tiamm iM^B i I — ^-^^ ^ r~ . .^- J. jjfc. f M IffiCMAL FRACTIONS. ^93; From th€ foregoing e^m|ires v^ may deduct thb followhig generiU To reduce a common to a deci$mU fircuxHon.'^ Annex oito o^ more ciphers, as may be necessary, to the numeratory and divide it by th^ denoihindtor. If then there be a romainder, annex another cijpher, and divide as before, and so eontinue to do, so long a^ there shall continue to be a remainder, or ufltil the fraction shall be. reduced to any necessary degree^ of exact- ness. The quotient will be the decimal required, which must con- sist of as many decimal places as thete are ciphers* annexed to the numerator ; and, if there are not so many figures in th^ quotient, the deficiency must be supplied by prefixing ciphers! BXAMPIiSS FOR PRACTICB. 4. Reduce |, i, 7^, and t^t ^ decimals. Ans. '5 ; '25 ; *025 ; *00797 -f% 5. Reduce |J, 1^, y^, and ^Vf to decimals. Ans, '692 + ; *003;/0028 + ; »000183+. 6. Reduce ^f , -^^^ vkhi to decimals. 7. Reduce |, ^, ^, J, |, A' A» ^ to decimals. 8. Reduce |, f , §, \, f , |, |, jt* A» A to decimals. Federal Moniey* IT 93* Federal Money is the currency cfthe United States The unit of English money is the pound sterling, which h divided into 20 equal parts, (twentieths,) called shillings; 3. What is the vulgar fraction equiyalent t» *7o4 7 Ans, i%i* 4. What i» the value of *6o3l — r- '014? *324 7 '6l02i 7 •2463 7 '6021 03 7 Ana. to last, sif^^^^* 5. What is the value of '437 * In this fraction, the repetend begins in the second place, or place of hun- dredths. The first figure, 4, is •^, and the repetend, 3, is ^ of ^, that i^ J^ ; these two parts must beadded tt^ether. ^ 4~ -^s-g^s^, Ans. Hence, to find the value of a mixied rqftetendi — Find the value of the two parts, separcUely, and add them tc^ether. 6. What is the value of '1637 iVi7+¥*ir=i^=^» ^^' 7. What is the value of '0047 7 Ana, y^^' 8. What is the va.ie of '1387 '16 7 '41237 It is plain, that circulates may be added, subtracted, multiplied, and di^ed, by first reducing them to their equivalent vulgar fractions. fas. DBOntAL VEAOnONS. 117 each dilliDg is divided into 12 parts called pence ; a penny beings 7^ of a pound. Each penny again is divided into 4 parts called farthings, a farthing being ^^ of a pound. These diTisions, therefore, are like those of common fractions, and the same difficoMes occur in operati<ms with English money as with common fractions. /The unit c^ Federal money is the Dollari divided into 10 parts called dttntf^frem a French word meaning tenth (of a dollar) ; each dime into 10 parts called cents, from the French for hundredik (of a dollar) ; and each cent into 10 parts called mtiilst from the French for thmuandth (of a dollar). These divisions of the money unit are like those of decimal fractions. Oar money, then, has this advantage over the English, viz., that operations in it are as in whole numbers, and we shall therefore consider it in connection with Decimals. The denominatioiJB of Federal money are eagles, dollars, dimes, cents, and miUs. TABUS. 10 miiis make 1 cent. 10 cents ( = 100 tniUs) 1 dime. 10 dimes (s 100 centf s 1000 miUs) 1 dollar. 10 dollars 1 eagle. Note. -^Coin is a jnece of metal stamped with certain impressions to give it a legal value, and also to serve as a guarantee for its weight and purity. The mill is so small that it is not usually regarded in busi- ness. The eagle is merely the name of a gold coin worth 10 dollars. Dimes are read as lOs of cents. Federal money, then, is calculated in dollars and cents, and accounts are kept in these denominations. A character, $, which may be regarded a contraction of U. S., placed before a number, signifies that it is Federal, or U. S. money. Questions. — IT 03. Wliat is Federal money? What is the unit of English money ? What are its denominations ? and what are they like ? What is the unit of Federal money ? how divided? and whence the names of the divisions ? What advantages has Federal over English money? Repeat the table. What is the ei^ ? How are diines read ? In what then is Federal money calculated, and accounts kept ? What b coin? What is the character for IT. S. money, and where placed? Where is the decimal point j^ed? Where and how many arc the ^aces for cents? for mills? why more places for cents than for mills? If the sum be but 8 cents how may it be written ? if three mills only, how ? How are 5 mills usually written ? 118 OBCI]|f!AL FRACTIOIfS. T94 As tbe doUar is the unit of Federal money, the decimal point is placed at the right hand of dollars ; and since dime^ (tenths) and cents (hundredths) are read together as cents the first two places at the right hand of the point express cents, and the third, mills, (thousandths.) Thus, 25 dollars 78 cents, and 6 mills, are written, $25*786. If there be no dimes, (tenths, oi\10s of cents,) that is, if the cents are less than 10, a cipher is put in ihe place of tenths ; thus, 8 cents are written 9*06 ; and if there are only mills, ciphers must be put in the place of tenths and hun- dredths ; thus, 5 mills are written 9*005.^ But 5 mills are usually expressed as half a cent; thus, 12 cents 5 mills, ars written 12^ cents, or $*12i. Reduction of Federal Money. IT 04* It is evident that dollars are reduced to cents in the same manner as whole numbers are reduced to kundredthSf by annexing two ciphers ; To TToUs or thousandths^ by annexing three ciphers. On the contrary, MiUs are redticed to cents by cutting off the right hand figure; To dollars, by cutting off three figures from the right, which is dividing by 1000, (T 41.) Cents are reduced to dollars by cutting off two figures from the right, which is dividing by 100. 1. Bednce Sd4 to cents. Am. 3400 cents. 3. Reduce $40'064 to mills. Ans. 40065 mills. 5. Reduce $16 to mills. Ans. 16000 mills. 7. Reduce $'75 to mills. Ans, 750 mills. 9. Reduce $*007 to mills. Ans. 7 mills. 2. Reduce 48143 mills to dollars. Ans. $48a43. 4. Reduce 48742 cents to dollars. Ans. $487'42. 6. Reduce 125 mills to cents. $*12|. 8. Reduce 2064} cents to dollars. Ans. $20'64j^. 10. Reduce 9 cents to dol^ lars. Ans. $'09. QaestioBfl. — If 94. How are dollars reduced to cents ? to mills ff cents to mills? mills to cents? to dollars? cents to dollars? T96. DECniAL FRACTIONS. 119 AddiUoii and Subtraction of Decimal Frac- tions. T AS* As the value of the decimal parts of a unit vary in a tenfola proportion like whole numbers, the addition and subtraction of decimal fractions, and of Federal money, may be performed as in whole numbers. 1. What is the amount 2. Frcmi 765*06 take 2t^ of 14*68, 9*045, 38^5, and <6805. 9K)025? Solution. — As nambers of the same denomination only csn be added together, (^ 12,) or subtracted from each other, the seyeral Bombers in each of these examples must be reduced to the lowest denomination contained in any one of the numbers, (% 91,) which is ten-thousandths, when the operation may be performed as in simple numbers. riRST 6pERATi(»ff. Or ifonly like denominations first operation. 14'6800 ^^'^ written under each other, 765*0600 9*0450 these alone will be added to- 27*6895 *Kii/zr\f\£\ gether, or subtracted from • Q«nn9? each other, and the opera- 737*3705, Am. yUU^o ^oos may be performed with- 71*2275 Ans, ^^^ ^^ formality of reducing to a common denom- ' * inator, since .the ciphers, by which the reduction is effected, make no difference in the result : thus, tBTjONO OPERATION. NoTE. — As the dcCi- SECOND OPERATION 14*68 mal point is at the right 765*06 9*045 of units, which are writ- 27*6895 ggtg ten under each other, the 9*0025 ^vrX in the result is di- 737*3705 rectly below the points in 71*2275 ^8 several numbers. Hence, To add or subtract decimal fractions^ Write the numbers under each other, tenths under tenths, hundredths under hundredths, &c., according to the value of their places ; add or subtract as in simple numbers, and point off in the result as many places for decimals as are equal to Questions* — If 95* Why can addition and subtraction of deci- mals be performed as in whole numbers? What numbers only can. be added and subtracted? How is this effected by the first operations? How by the second? How do you write down decimals for addition? How for mbtraction ? Where place the point in the results? Repeat the rule How do you prove addition of aecimals ? How subtraction t 120 DBCUfAL FRACTIONa %9S^ the gre^xeA nmober of decimal places in any of the giveo. numbers. Proof. — The same as in the addidon and aabtraction of ain^pto numbers KXAMDPIiBS FOR PRACTICIB. 3. A man sold wheat at several times as follows, Tiz.» 13^ bushels, 8*4 bushels, 23'051 bushels^e bushels, and *7S of a bushel ; how much did he sell in the whole ? Am. 5V451 bushels. 4. What is the amount of 4^, 21^, 355^^, lyj^, and l^ ? Am. SOS^Vife, or 808*143. 6. What is the amount of 2 tenths, 80 hundredths, 89 thou<^ sandths, 6 thousandths, 9 tenths, and 6 thousandths? Ans. 2. 6. What is the amount of three hundred twenty-nine and seven tenths, thirty-seven and one hundred aixty>4wo tkou* sandths, and sixteen hundredths ? 7. From thirty-five thousand take thirty-five thousandths. Am. 34999*965. 8. From 6*83 take 4*2793. Am. 1*5507. 9. From 480 take 245*0075. Am.2U9925. 10. What is the difierence between 1793*13 and 817*05 693 ? Am. 976*07307. 11. From 4^ take 2^^^. Remainder, 1^, or 1*98. 12. What is the amount of 29^, 374ny,,#,^, 97t^^. 315^^, 27, and 100^ ? Am. 942*957009. Examples in Federal Mtmey can emdenthf he performed in the same way. 1. Bought 1 barrel of flour for 6 dollars 75 cents, 10 pounds of cofiee for 2 dollars 30 cents, 7 pounds of sugar for 92 cents, 1 pound of raisins for 12^ cents, and 2 oranges for 6 cents - what was the whole amount? Ans. $10*155. 2. A man is indebted to A, $237*62; to B, $350; to C, $86a2i ; to D, $9*62| ; and to E, $0*834; what is the amount of his debts ? Am. $684*204. 3. A man has three notes specifying the following sums, viz., three hundred dollars, fifty dollars sixty cents, and nine dollars eight cents ; what is the amount of the three notes? Am. $359*6a 4. What is the amount of $66a8, $7<37i, $280, $0*287 $17, and $90*413 ? Am. $451*265. n f 96. DBCniAL FRACTIONa 121 5. Bought a pair of oxen for S76'50, a horse for 985, and a cow for 817*25 ; what was the whole amount ? Am, $178*75. 6. Bought a gallon of molasses for 28 cents, a quarter of tea for 37^ C3nts, a pound of saltpetre for 24 cents, 2 yards of broadcloth for 11 dollars, 7 yards of flannel for 1 dollar 624 cents, a skein of silk for 6 cents, and a «tick of twist for I cents; how much for the whole? Am. $13*62. 7. A man bought a cow for eighteen dollars, and sold her again for twenty-one dollars thirty-seven and a half cents ; how much did he gain ? Am. $3*375. 8. A man hought a horse for 82 dollars, and sold him again for seventy-nine dollars seventy-five cents ; did he gain or lose ? and how much ? Am. He lost $2*25. 9. A merchant bought a piece u)f cloth for $176, which proving to have been damaged, he is willing to lose on it •16*50 ; what must he have for it ? Am. $159*50. 10. A man sold a farm for $5400, which was $725*37^ niore than he gave for it ; what did he give for the farm ? 11. A man, having $500 dollars, lost 83 cents; how much Had he left ? Am. $499*17. 12. A man's income is $1200 a year, and he spends S800*35 ; how much does he lay up ? 13. Subtract half a cent from seven doUars. Rem. $6*99J. 14. How much must you add to $16*82 to make $25 ? 15. How much must you subtract from $250, to leave 187*14? 16. A man bought a barrel of flour for $6*25, 7 pounds of coffee for $1*41, he paid a ten dollar bill ; how much must he receive back in change ? Am. $2*34. MnItipUcation of Decimal Fractions. IT 96. 1. Multiply *7 by '3. , *7 = ^ and *3 = T%, then ^X^ = ^ = ^2\, Am. We here see that tenths multiplied into tenths produce hun- dredths, just as tens, (70,) into tens, (30,) make hundreds, (2100.) We may write down the numerators decimally, thus: 11 122 raiClMAL FRACTIONS. 1[96. QPKRATION. •7 -!? *21, Ans. The 21 m(U9t be htindredths as before. The numbef of figures in the product, it will be seen, is equa] to the / number in the multiplicand and multiplier ; hence, we have as many places for decimals in the product 48 there *- are in hoih. the factors. •3 ■" Note. — The correctness of the above rulfe may be illustrated by the annexed diagram. The length of the plot of ground which it represent may be regarded 10 feet and the breadth 10 feet, each division of a line, consequently, being OQe tenth of the whole line. Multiplying 10 by 10, we have 100 square feet in the plot, each of the small squares being 1 square foot, or one hun- dredth of the whole plot. Now take the part encircled by tlie black lines, 7 feet ('7 of the whole line) long, and 3 feet (*3 of the whole line) wide. The contents are 91 square feet, or *21 of the whole plot ; hence the product of '7 into *3 is *21 as above. 2. Multiply 425 by »03. Here, as the number of significant figures in the product is not equal to the number of decimals in both factors, the deficiency must be supplied by pre- fixing ciphers, that is, placing them at the left hand. The correctness of the rule may appear from the following process : *125 is t^^j^j ^^^ *^3 ^ tSt> • tWtt X T*Tr = T^^Ais — *00375, the OPERATION. a25 *03 »00375 Prod. now, same as before. Hence, To multiply decimal fractiom, RULE. Multiply as in whole numbers, and from the right hand of the product point off as many figures for decimals as there are decimal places in the multiplicand and multiplier counted together, and if there are not so many figures in the product, supply the deficiency by prefixing ciphers. Questions. — T[ 96* Tenths X tenths produce what ? Illustrate this by a diagram. To what must the number of places in the product be equal ? When the number of decimals in the product is less than the number in both factors what do you do ? How can you tell of what nanie or denomination will be the product of one given decimal mul- tiplied into another given decimal, without going through the process of multiplication ? Of what denomination, then, will be the product of •leX'ZS? of*0005X*07? Td7. OBOOfAl. PRACTiOWi U| MSLAMPfJm FOR FBACTICBi 3. Multiply five hundredths by seven thousandths. • Product, 'OOOSS. 4. What is *3 of 116 ? ^w*. 34*8. 5. What is *85 of 3672 ? Am. S\2h2. 6. What is *37 «f K)663 ? i!7w. •020631. ^ 7. Multiply 672 by '58. Product, 331*76. 8. Multiply eighty-six by four hundredths. Product, 3*44. 9. Multiply ^0062 by •OOOS. 10. Multiply forty-seven tenths by one thousand eighty-six hundredths. Prod. 51*042. HXAMPUBS IN FEPBRAIi MOHnCT. fl" 97. 1. If a melon be worth $*09, what is *7 of it worth? (1F77.) Am, $*063. 2. What will 250 bushels of rye cost, at $'88^ per bushel ? 3. What is the value of 87 barrels of flour, at $6*37J a barrel ? Am, $554*62 J. 4. What will be the cost of a hogshead of molasses, con- taining 63 gallons, at 28| cents a gallon ? Am, $17*955. 5. If a man spend 12| cents a day, what will that amount to in a year of 365 days ? What will it amount to in 5 years? Am. It will amount to $228* 12 J in 5 years. 6. If it cost $36*75 to clothe a soldier 1 year, how much will it cost to c'.othe an army of 17800 men ? Am. $654150. 7. Multiply $367 by 46. 8. Multiply $0*273 by 8600. iln*. $2347*80. 9. At $5*47 per yard, what cost 8*3 yards of cloth. ? Am. $45*401 10. At $'07 per pound, what cost 86*5 pounds of rice ? Am. $1*855. 11. What will be the cost of thirteen hundredths of a ton of hay, at $11 a ton ? - Am. $1*43. 12. What will be the cost of three hundred seventy-five thousandths of a cord of wood, at $2 a cord ? $*75. 13. If a man's wages be seventy-five hundredths of a dol- lar a day, how much will he earn in 4 wee&s, Sundays excepted ? Am. $18. 124 DECIMAL FRACTIONS. T98. Division of Decimal Fractions. f 98. 1. Divide *21 by '3. *21=^, and *3==tSV- Now 1%^-^^; or ^ of ^ = f^=-ft. It appears, then, that hundredths divided by tenths give tenths, just as hundreds (2100) divided by tens (30) gvm tens, (70.) The numerators may be set down decimally, and the division per- formed as follows ; — OFESATION. The 7 must be tenths as before. The dividend , *3)^21 which answers to the product in multipUcation, con- tains two decimal places ; and the divisor and quotient, *7 which answer to the factors in multipUcation, (U^ 31,) together contain two decimal places. Hence, we see that the number of decimal places in the quotient is equal to the dif ference between the number in the dividend and divisor. 2. At 4*75 of a dollar per barrel, how many barrels of flour can be bought for $31 ? OPERATION. The 4*75 are 475 hundredths, and, since 4*75) 31*00 (6'5264- the dividend and divisor must be of the 2850 same denomination, we annex 2 ciphers to 31 and it becomes 3100 hundredths, (If 91 .) 2500 Then there can be as many whole barrels 2375 bought as the number of times 475 bun- — dredths can be subtracted from 3100 hun- 1250 dredths. The 6 barrels thus found will 950 cost 2850 hundredths of a dollar, and as ~^77T" 250 hundreths or cents remain, it will buy OQ^ part of another barrel, which we find by 2850 annexing ciphers, and continuing the op- 150 eration. We now see that there are 5 decimal places in the dividend, counting all the ciphers that are annexed, and as there are but two in the divisor, we point off 3 in the quotient. There is still a remainder of 150, which, written over the divisor, (Tf 36,) gives Iff of a thousandth of a barrel, a quantity so small that it may be neglected. But we place -f- at the right of the last quotient figure, to show that there is more fiour than indicated by the quotient. Ans, 6*526+ barrels. Note. — It is sufilciently exact for most practical purposes to carry tlie division to three decimal places. 3. Divide *00375 by * 125. OPERATION. The divisor, 125, in 375, goes 3 times, and 125 ) *00375 ( *03 no remainder. We have only to place the deci- 375 mal point in the quotient, and the work is done. There are five decimal places in the dividend ; 000 •onsequently there must be five in the divisor rOa DECIMAL FRAGTIONa 120 and quotient eonnted together ; and as there are three in the dlTiacnr, there mast be two in the quotient ; and, mnce we hare but one figure in the quotient, the deficiency roust be suppUed by prefixing a cipher. The operation b^^ vulgar fractions will bring us to the same result. Thus, 425 is t\^, and *00375 is ttAWW • now, TT«tftnF ^ iWtf = tH^Mtf = t#tf = *03, the same as before. 4. Divide «76 by •005. OPERATION. Solution. — We cannot divide hundredths (75 by thousandths, until the former are reduced to ]^Q thousandths by multiplying by 10, or annexing one cipher, when the mviBor and dividend wiU K)05 ) *750 be of the same denomination ; and '005 is con- "~" tained in (can be subtracted from) '750, 150 150, Quot. tiines, the quotient being a whole number. These illustrations will establish the following I. Reduce, if necessary, the dividend to the lowest denom- ination in the divisor, divide as in whole numbers, annexing ciphers to a remainder which may occur, and continuing the operation II. If the decimal places in the dividend with the ciphers annexed exceed those in the divisor, point off the excess from I the right of the quotient as decimals ; but if the excess is more than the number of places in the quotient, supply the deficiency by prefixing ciphers. EXAMPLES FOR PRACTICE. 5. Divide 3156'293 by 2547. Qwo^. 125*34-. 6. Divide 173948 by '375. Quot. 463861 -f. Note. — The pupil will point off the decimal places in the quo- tient of this and the following example, as directed by the rule. 7. Divide 5737 by 13'3. Quot. 431353. 8. What is the quotient of 2464'8 divided by '008 ? • Ans. 308100. Qnestions. — IT 98. Hundredths, divided by tenths, give what ? How is it in integers ? Exhibit on the blackboard the process of divid- iDg 7 by 1 '25. Why do you annex ciphers to the 7 ? What is the quo- tient? Why pointed thus ? Give a demonstration by common fractions, as after Ex. 3, and show that this placing of the point is right. The sign of addition, annexed to the quotient, is an indication of what ? When there are remainders, to how many places should the division b« carried? Why not to more places? Repeat the rule for division 11# iS6 DfiCIMAL FBAGTIONS. V99 9. Divide 2 by (^a. Qteof. *037+. 10. Divide *012 by '005, Quot. 2*4. 11. Dividie three thousandths by four hyindredths. QuoU *075. 12. Divide eighty-six tenths by ninety-four thousandths. 13. How many times is *17 contained in 8? BXAMPIiES IN FSDKRAIi MONEY. IT 99. 1. Divide $59*387 equally among 8 men; how much will each man receive ? OPERATION. 8) 59*387 Am, f7*423f , that is, 7 dollars, 42 cents, 3 mills, and f of another mill. The | is the remainder, after the last division, written over the divisor, and expresses such fractional part of another mill. For most purposes of business, it wiU be sufficiently exact to carry the quotient only to mills, as the parts of a mill are of so little value as to be disregarded. 2. At $*75 per bushel, how many bushels of rye can be bought for $141 ? Am, 188 bushels. 3. At 12 J cents per lb., how many pounds of butter may be bought for $37 ? Am. 296 lbs. 4. At 6^ cents apiece, how many oranges may be bought for $8 ? Am, 128 oranges. 5. If *6 of a barrel of flour cost $5, what is that per bar rel? / -. Am, $8'333+. Note. — If the sum to be divided contain only dollars, or dollan and cents, it may be reduced to mills, by annexing eiphers before dividing ; or, we may first divide, annexing ciphers to the remainder, if there shall be any, till it shall be ledu^ to mills, and the result will be the same. 6. If I pay $468*75 for 750 pounds of wool, what is the value of 1 pound ? Am, $0'625 ; or thus, $*62J. 7. If a piece of cloth, measuring 125 yards, cost $181'25, what is that a yard ? Arts, $1'45. 8.- If 536 quintals of fish cost $1913*52, how much is that d quintal ? Am, $3'57 9. Bought a farm, containing 84 acres, for $3213*; what did it cost me per acre ? Ans, $38*25. 10. At $954 for 3816 yards of flannel, what is that a yard I Am, $0*25. r s f 106. DECIMAL FRACTIONS. ISf IL Bought 72 pounds of raisins for 88; what was that a Eoand? Ans. $0*11U; or, $0*111+. ^12. Divide $12 into 200 equal parts ; now much is (me of the parts ? ^^^ = how much ? Am, 8*06. 13. Divide $30 by 750. ^ = how much ? 14. Divide $60 by 1200. yj^ = how much ? 15. Divide $215 into 86 equal parts ; how much will on^ of the parts be ? ^^ = how much ? IT lOO. Review of Decimal Fractions. Questions* — What are decimal fractions? How do they differ from common fractions ? How can the proper denominator to a decimal fraction be known, if it be not expressed ? What advantages have deci- mal over common fractions ? How is the value of every figure deter- Dttined? Describe the manner of numerating and reading decimal frac- tions ? of writing them ? How are decimals, having different denomi- nators, reduced to a common denominator? How may any whole namber be reduced to decimal parts ? How can any mixed number be read together, and the whole expressed in the form of a common frac- tion? What is federal mtmey ? What is the money unit, and what are its divisions and subdivisions ? How is a common fraction reduced to a decimal ? To what do the denominations of federal money correspond I What is the rule for addition and subtraction of decimals? — multiplica- tion ? — division ? KXBRCISES. * 1. A merchant had several remnants of cloth, measuring as follows, viz. : 7 S vds. "J How many yards in the whole, and what would the whole come to, at $3'67 per yard ? Note. ^-Reduce the conmion fractions to decimals. Do the same wherever they occur in the examples which follow. Ans. 36*475 yards. $133'863 +, cost. 2. From a piece of cloth, containing 36§ yards, a merchant sold, at one time, 7^ yards, and, at another time, 12f yards ; how much of the cloth had he left? Ans, 16'7 yds. 3. A farmer bought 7 yards of broadcloth for $33f J, two barrels of flour for $14,^, three casks of lime for $7|., and 7 pounds of rice for $f ; what was the cost of the whole ? Note*. — The following examples are to be performed accoroing to the rule in ^77, or in 1 83, or in 1 85. 4 At 124 cents per lb., what will 37| lbs. of butter cost? Ans. $4*718i. i 128 DECIMAL FRACTIONS. IT 100. 5. At $17'37 per ton for hay, what will llf tons cost? Am. S20r92f. 6. The above example reversed. At $201*92f for llf tona of hay, what is that per ton ? Am. $17*37. 7. If »45 of a ton of hay cost S9, what is that per ton ? Am. $20. 8. At *4 of a dollar a gallon, what will *25 of a gallon of molasses cost? Ans. $'1. 9. What will 2300 lbs. of hay come to, at 7 mills per lb. ? Am. $1640. 10. What will 765J lbs. of coffee come to; at 18 cents per lb. ? A?ts. $137'79. 11. Bought 23 firkins of butter, each containing 42 pounds, for 16 J cents a pound ; what would that be a firkin ? and how much for the whole ? Ans. $159'39 for the whole. 12. A man killed a beef, which he sold as follows, viz., the hind quarters, weighing 129 pounds each^ for 5 cents a pound ; the fore quarters, one weighing 123 pounds, and the other 125 pounds, for 4J cents a pound ; the hide and tallow, weighing 163 pounds, for 7 cents a pound ; to what did the whole amount ? Am. $35*47. 13. A farmer bought 25 pounds of clover seed at 11 cents a pound, 3 pecks of herds grass seed for $2*25, a barrel of flour for $6*50, 13 poujids of sugar at 12 J cents a pound ; for which he paid 3 cheeses, each weighing 2i7 pounds, at 8 J cents a pound, and 5 barrels of cider at $1*25 a barrel. The bal- ance between the articles bought and sold is 1 cent; is it^br or against the farmer ? 14. A man dies, leaving an estate of $71600 ; there are demands against the estate, amounting to $39876*74; the residue is to be divided between 7 sons ; what will each one receive ? Am. $4531'894f . 15. How much coffee, at 25 cents a pound, may be had for 100 bushels of rye, at 87 cents a bushel ? Am. 348 pounds. 16. At 12J cents a pound, what must be paid for 3 boxes of sugar, each containing 126 pounds ? Am. $47*25. 17.. If 650 men receive $86.*75 each, what will they all receive ? Am. $56387*50. 18. A merchant sold 275 pounds of iron at 61 cents a pound, and took his pay in oats, at $0*50 a bushel; how many bushels did he receive ? Am. 34*375 bushels. 19. How many yards of cloth, at $4*66 a yard, must be given for 18 barrels of flour, at $9*32 a barrel ? Am, 36 yards* r flOl. BILLa 129 20. What is the price of three pieces of cloth, the first con- taining 16 yards, at 83*75 a yard ; the second, 21 yards, at $4*50 a yard ; and the third, 35 yards, at 85* 12^ a yard ? Am. 8333^87^. BILLS. IT lOl • A Bill, in husiness transactions, is a written lis! ot the articles bought or sold, and their prices, together with the entire cost or amount cast up. No. 1,^^ BUI of Sale, Payment received, Boston, May 25th, 1847. James Brown, Esq. Bought of Hastings &; Belding, 6 yards black broadcloth, 83*00 2i « cambric, " *14 2 dozen buttons, " *15 4 skeins sewing silk, " *04 25 lbs. brown sugar, " '09 Received payment, $21*06. Hastings 6c Belding. JVb. 2. — Bill of Sale, Charged in account. New Orleans, Aug. 1st, 1847. Gren. Z. Taylor, To Daniels &; Thomas, Dr. To 278 bbls. beef, ** 191 " pork, " 250 ** flour, ** 500 sacks Indian meal. B 89*75 i( 12*00 <( 5*70 C( *62 Charged in acc't. Amount, $6741*25 Daniels & Thomas. No. 3.— Barter Bill. Buffalo, Sept. 15th, 1847. Mr. D. F. Standart, To O. B. Hopkins & Co., Dr. To 15 lbs. brown sugar, • $ *10 ** 2 ** Y. H. tea, « *87; •* 24 *« mackerel, " '04] " 3 gal. molasses, " *42 ** 16 yds. sheeting, ** * '09 T 190 BiLLa Y MH Gr. By 4 doz. eggs, m »<08 " 8 lbs. butter, c< 44 " 40 " cheese, M *07i " note at 30 days, to balance. 2*59 87*03 O. B. Ho^ins & Go. by L. D. Swift. No. 4. — Bin of goods sold at wholesale. New-Yorki April 5th, 1847. Davis & Horton, Bought of Barnes Porter & Go. 3 hhds. molasses, 118 gal. eaca, B $ *31 2 « brown sugar, 975 aud 860 lbs. " *09J 3 casks rice, 205 lbs. each, «♦ *04| 5 sacks coffee, 75 " ' " *11 1 chest H. tea, 86 " *• " *92 $43M6 Bec'd payment, by note, at 60 days^ For Barnes rorter & Co. James D. Willard. It is sometimes practised, in collecting and settling accounts, to make a copy of each individual account, and present it to the person for his inspection. No. 5. — Copy of an individzuil account. Frank H. Wright, In acc't with Edward F. Gooper^ 1847. Dr. Jan. 7. To 125 busheb com, J» $ *50 « " " 20 " apples, " *31 March 13. " 12 " rye, « *62 " 20, « 15| lbs. cast stefeU " *24 Question s. ^- If 1 1 • What is a bill ? If the amount of the bill be paid at the time, how is it shown? Which bill is an example of this? if charged in account, how is it shown? example ? How does a baiter bill dififer from a bill of sale ? In what order are the articles bought and sold arranged ? What is practised in collecting and settling accounts? How does such a copy difier from a barter bill ? To which of the biUft most the biU to be made out conform ? and what will it be c^ed ? Tlt2. COifPOUND NUMBBRa 181 1847. Cr. Feb. 15. By 3 cows, m tl7<00 " 22. '• 5 sheep, « 2*60 Baltimore, May 9th, 1847. Amount due me, S16,42 Edward F. Cooper. The pupil is required to make out a bill from the statement contained in the following example. Wm. Prentiss sold to David S. Piatt 780 lbs. of pork, at 6 cents per lb. ; 250 lbs. of cheese, at 8 cents per lb. ; and 154 lbs. of butter, at 15 cents per lb. ; in pay he received 60 lbs. of sugar, at 10 cents per lb. ; 15 gallons of molasses, at 42 cents per gallon ; J barrel of mackerel, $3*75 ; 4 bushels of salt, at $1*25 per bushel ; and the balance in money : how much money did he receive ? Ans. $68*85. COMPOUND NUMBERS. if 108. When several abstract numbers, or several de- nominate numbers of the same unit value, are employed in an arithmetical calculation, they are called simple numbers^ and operations with such numbers are called operations in simple numbers. Thus, if it were required to add together 7 gallons, 9 gallons, and 5 gallons, the numbers are simple numbers, being denominate numbers of the same unit value, (1 gal.,) and the operation is an addition of simple numbers. We have had, also, subtraction, multiplication, and division ot simple numbers. But when several numbers of different unit values are em- ployed to express one quantity, the whole together is called a compound number. Thus, 12 rods, 9 yards, 2 feet, 6 inches, 'employed to express the length of a field, is a compound num- ber. So also, 9 gallons, 2 quarts, 1 pint, employed to express a quantitj'^ of water, is a compound number. Note. — The word denomination is used in compound numbers to Questions. — If 102. What are simple numbers? Examples. What are operations in such numbers called ? What is a compound number ? Give examples other than those in the book. Wlat is meant by the word denomination ? ^ 132 COMPOUND NUMBERS. IF 103, 104. denote the name of the unit considered. Thus, bushel and peck aura names or denominations of measure ; hour^ minute and second aie do- nominations of time. IT 103. The fundamental operations of addition, subtrac ' tion, multiplication and division, cannot be performed on com- pound numbers till we are acquainted with the method of changing numbers of one denomination to another without altering their value, which is called Reduction, Thus, we wish to add 2 bushels 3 pecks, and 3 bushels 1 peck, together. They will not make 9 bushels nor 9 pecks, (adding together the several numbers,) since some of the numbers express bushels, and some express pecks. But 2 bushels equal 8 pecks, (2 times 4 pecks, the num)?er of pecks in a bushel,) and 3 pecks added make 11 pecks ; 3 bushels equal 12 pecks, and 1 peck added make 13 pecks. Then, 11 pecks -|- 13 pecks = 24 pecks. Hence, before proceeding further, we must attend to the Reduction of Compound Numbers. Sterling or English Money. IT 104. Money is expressed in different denominations, and 4 dollars, 3 dimes, 7 cents, 5 mills = $4'375, employed to express one sum in Federal money is a compound number. But as the denominations in Federal money vary unifornily in a tenfold proportion, (IT 93,) being conformed to the Arabic notation of whole numbers, the operations in it are as in whole numbers. The denominations in English (called, also, sterling) money^ pounds, shillings, pence and farthings, do not vary uniformly, but according to the following Note 1. — All the tables in Reduction of Compound Numbera must be carefully committed to memory by the pupil. 4 farthings (qrs.) make 1 penny, marked d, 12 pence (plural of penny) 1 shilling, " s, 20 shillings 1 pound, " £. Note 2. — Farthings are often written as the fraction of a^nny, thus, 1 farthing = id., 2 farthings = id., 3 farthings = |d. Questions. — If 103. What is reduction ? Whence its necessity f Explain by the example of adding bushels and pecks. To what, then, mast we attend before proceeding further? T104. COMPOUND NUMBERS. 133 NoTc 3. — The raloe of these denominatiooa in Federal money is Heady as fdlows : Iqr. = Hi^f^ <^*^- Id. = 2^ ceTtts, Is, = 24j cents. 1£. = S4*84 4y. Id, 2^^^*. ==$1»00 1. How many farthings in 5 pence ? 1st operation. Solution. — There is in Enriand a gold coin, called a sovereign, the Tnlue of which is J^l. We may multi- ply the nomher of farthings (4) in 1 penny by the number of pence, (5.) (Tf 46.) Or, as ei- ther factor may be made the multiplicand, {% 21,) we may multiply the number of pence (5) by the number of farthings in 1 penny. Ans, 20qrs. 3. How many farthings in 3 pence ? 6 peace ? — 9 pence ? 7 pence ? 2 pence 1 10 pence ? 4 5 20qrs, 2d operation. 5 4 20^*. 12 pence: 11 pence ? ■ 1 shilling? 5. How many pence in 3 shillings ? 5 shillings ? 5s. 8d.? 7s.? 8s. 4d. ? 12s. ? 15s. 6d. ? 7. How many shillings in £3? £5? £4 2s.? £6 lis.? 2. In 20 farthings, how many pence ? « Solution. -^ We operation, have given the num- 4 ) 20 her of farthings in 1 — penny to find the 5^, number of pence in a given number of far- things, (20,) and we divide the number of farthings in the num- ber of pence by the number in 1 penny, (Tf 46.) Ans. 5d. 4. How many pence in 12 farthings? 24 farthings? 36 farthings ? 28 farthings ? 8 farthings ? 40 farthings ? 44 farthings ? 48 farthings ? 6. How many shillings in 36 pence ? 60 pence ? 68d. ? 84d. ? lOOd.? 144d.? 186d. ? 8. How many pounds in 60s.? lOOs.? 82s.? >r^ 131s. ? Questions. — IF 104. WTiat is said of operations in Federal money ? Wha. are the denominations of English money? the signs? How do they vary differently from those of Federal money? Give the table. How are farthings written ? What is the value of a pound sterhng in Federal money ? Explain the first operation of Ex. 1 : the second oper- ation Explain Ex. 2. Of how many kinds is reduction ? what ars they What is reduction descending ? — reduction ascending ? 12 134 COMPOUND NUMBERS. T109. The changing of higher denominations to lower, as pounds to shillings, is called Reduction Descending^ and is performed by multiplication. REDUCTION DESCENDING. IT 105. 1. In £17 13s. 6|d., how many farthings ? OPERATION. 17£ 135. M, 2qrs. 20 353*. in 17£ 13*. 12 4242<i. in 17£ 13*. W. 4 1697l9r*. Am, In 17£ 13s. M. 2qr8, Solution. — We multiply VI £ by 20, the shillings in iJb, and add in tiie 13s. to get the number of shillings in 17£ 13s., which is 353. This number we multiply by 12, adding in the 6d. given, to get the number of pence, 4242, which we multiply by 4, adding in the 3qr8. given, to get {he num- ber of qrs. or farthings, which \a 1697lqrs. Hence, for Redtiction Be- scendingf RULiE* Multiply each higher de- nomination by the number which it takes of the next less The changing of lower de- nominations to higher, as shil<^ lings to pounds, is called iZe- dtiction Ascending^ and is performed by division. REDUCTION ASCENDING. 2. In 16971 farthings, how many pounds ? OPERATION. ^^TSS;g;'*}4) 16971 i2SS.J;|12)4242d3gr* •ii'!25:12|0)35|35. 6i. 17£ 13*. Ans. £17 135. ^, 3^*. Solution. — We divide the whole number of farthings by 4, the number in Id., to get the number of pence ; for as many times as 4 can he subtracted from 16971, so many pence there will be, which is 4242d. and 3qrs. re- maining. On the same principle, dividing the 4242 by 12, the quo- tient, 353, is shillings, and the remainder, 6, is pence, and divid- ing 353s. by 20, the quotient, 17, is pounds, and the remainder, 13, i&\ shillings. Hence, for Reduction As-^ cending, RUUB. Divide each lower denomi- nation by the number which it takes of it to make one of Questions* — ^ 105* Explain the first example. Give the rule ior re<lucti< n descending. Ex. 2. Give the rule for reduction ascending. / TlOft. coMPOinfD ifUMinw. 136 to mftke 1 of thk higher, in* the next hiffher. Proeeecl in creasing the product by the this way till the work is done. given number, if any, of this lower denomination. Proceed in this way till the work is done. BXAMPLiES FOR PRACTICE. 3. Reduce £32 15s. 8d. to 4 Reduce 31472 farthings qrs. to pounds. 5. Reduce £7 148. 6d. 1 6. Reduce 7417 qrs to qr. to qrs. pounds. 7. In £91 lis. 3jd., how 8. In 87902 farthings, now many farthings ? many pounds ? 9. In £40 128. 8d., how, 10. In 9752 pence, how many pence ? many pounds ? 11. In £1 188. 4|d., how 12. In 921 half pence, how many half pence ? many pounds ? Weight. I. AvoiEDUPois Weight. T 106. Avoirdupois Weight is employed in all the ordi- nary purposes of weighing. The denominations are tons^ pounds, ounces, and drams. TABIiB. 16 drams (drs.) make 1 ounce, 16 ounces " 1 pound, 2000 pounds " 1 ton, Or, as wax formerly reckonedy 28 lbs. 1 quarter, 4 qrs. {=112 lbs.) 1 hundred weight, 20 cwt. (=2240 lbs.) 1 ton. By the last taHe, 2240 lbs. make 1 ton, which is sometimes called the " long ton;" while the ton of 2000 lbs. is. called the " short ton." The long ton is still used in the U. S. cus- Qnestions. — IT 106. What is the use of avoirdupois weight ? the denominations? the signs? Repeat the table; the table by the old method. Explain the difference between the long and short ton. When f 8 the long ten used ? the short ton ? marked oz. it lb. (C T. (« qr. , «• cwt. (( T. / V Ida COMPOUND NUMBERa IT 107 tom-house operations, in invoices of English goods, and of coal from the Pennsylvania mines. But in selling coal in cities, and in other transactions, unless otherwise stipulated, 2000 lbs. are called a ton. BXAMPLiES FOR PRACTICE 1. In 14 tons 607 lbs. 6 oz. 2. In 7323500 drams, how 12 drs., how many drams ? many tons ? Solution. — ^Dividing the drams by 16, the number in an oz., the quotient is oz. and the remainder drs. Dividing the oz. by 16, the quotient is lbs. and the remainder oz., and dividing the lbs. -by 2000, the quotient is tons and the re- mainder lbs. 4. In 14665 lbs. of sugar, how many tons ? 6. In 7323500 drams, how many tons ? 8. In 470 packages of screws, each containing 26 lbs., how many tons ? Solution. — As there are 2000 lbs. hi a ton, we multiply 14 by 2000, to get 14 tons to lbs., and add the 607 lbs. to the product. The lbs. we multiply by 16 to get them to oz., adding in 6 oz., and the oz. by 16, adding in 12, and the whole are in drams. 3. In 7 tons 665 lbs. of su- gar, how many lbs. ? 5. In 12 T. 15 cwt. 1 qr. 19 lbs. 6 oz. 12 drs. of glass, re- ceived from an English house, how many drams ? 7. Received from Birming- ham, England, 5 T. 9 cwt. 12 lbs. of iron screws, in pack- ages of 26 lbs. each ; how many were the packages? ' ^ /. ! II. Troy Weight. IT lOT. Troy Weight is used where great accuracy is required, as in weighing gold, silver, and jewels. The de- nominations are pounds, ounces, pennyweights, and grains. TABLE. 24 grains (grs.) make 1 pennjrweight, marked pwt. 20 pwts. 1 ounce, " oz. 12 oz. 1 pound, " lb. Note. — A lb. Troy = 5760 grs., and 1 lb. avoirdupois = 7000 grs. Troy. Hence a quantity expressed in one weight, may be changed to the denominations of the other. Questions. — IF 107. For what is Troy weight used ? What are the denominations ? — the signs ? Repeal the table. What difference between the pound Troy and the pound avoirdupois ? r fioa COMPOUND NUMBERS. 137 BXAMPLiES FOR PRACTICB. 1. In 210 lbs. 8 oz. 12 pwts., how many pwts. ? Solution. — Multiply the Ibe. by 12, adding the '8 oz. to the product, and the sum is oz., which, multiplying by 20, adding in the 12 pwts., the sum is pwts. 3. In 7 lbs. 11 oz. 3 pwts. 9 grs. of silver, how many grains? 5. Reduce 11 oz. 13 pwts. 13 grs. of gold to grains. 7. Reduce 28 lbs. avoirdu- pois to the denominations of Troy weight ? 2. In 50572 pwts., how many lbs. ? Solution. — Dividing the pwts. by 20, the quotient is oz., and the remainder pwts., and dividing the oz. by 12, the quotient is lbs., and the remainder oz. 4. In 45681 grains of sil- ver, how many lbs. ? 6. Reduce 5605 grs. of gold to ounces. 8. Reduce 34 lbs. 6 pwts. 16 grs. Troy to lbs. avoirdu- pois. (Consult Note.) ni. Apothecaries' Weight. IT 108. Apothecaries' Weight is used by apothecanes and physicians, in mixing and preparing medicines. But medicines are bought and sold by avoirdupois weight. The denominations are pounds, ounces, drams, scruples, and grains* 20 grains (grs.) make 1 scruple, marked 9. SB 1 dram, " 5. 8 3 1 ounce, " g. 12 g 1 pound, " ft. Note. — The pound and ounce. Apothecaries' and Troy weight, are the same, but the ounce is differently divided. EXAMPLiES FOR PRACTICK* 1. In91b.8g. 15.2 9. 19 grs. how many grains ^ lbs 2. Reduce 55799 grs. to Questions. — TF 108. To what is apothecaries' weight limited! the denominations ? Give the table. Make the sign for each denomi- muion. What is S9id of the pound and ounce? 12=**= 138 COMPOUND m7SI]^BS. T 109. Measures of Extension. Extension has three dimensions, length, hreadth, and thick- ness. I. Linear Measubi^ T !©©• Linear Measure (the measure of lines) is used when only one dimension is considered, which may be either the length, breadth, or thickness. * The usual denominations are miles, furlongs, rods, yards, feet, inches, and barley-corns. TABUB. 3 barley-corns (bar.) make 1 inch, marked in. 12 inches 1 foot, 3 ft. 1 yard, 5J yards, or 16J ft, 1 rod, 40 rods 1 furlong, 8 furlongs, or 320 rods, 1 mile, ft. yd. rd. fur. mi. 69^ common miles, 1 degree, deg., or **,|'*'i^^%?Sf;Sh.'^""^ 3 geographical miles, 1 league, L., JuMdiiimea«uringd«t*nceiiatMa. 60 geographical miles, 1 degree of latitude. 6 feet, 1 fathom, in measuring depths at sea. NoTB. — The geographical mile, used in measuring latitude, is not quite uniform, but is always a little less than 1^ comimon miles, as Uie degree varies from 68| to 69| miles. The degree of longitude grows shorter towards the poles, where it is nothing. Instead of the barley-corn, inches are now divided into eighths and tenths. ESXAMPLSS FOR PRACTICE. 1. How many inches in the 2. In 1577664000 inches, equatorial circumference of how many miles ? How many the earth, it being 360 de- degrees of the equatorial cir- grees ? cumference ? Questions. — ^ 109. How many dimensions has extension? What are they ? What is Unear measure ? Give the denominations, and the sign of each. Repeat the table. How is the inch usually divided ? For what is the fathom used ? For what is the geographical mile used ? — its length ? What is said of the degree of longitude, and its length ? — of a degree of latitude ? What causes the difference in the length of degrees? Fcr what is the league used, and about what is its ^ength in common mdlej? r tllO. COIIPOUlfD IfUMfiKS. 199 3. How many inches from Boston to Waslungton, it be- ing 482 miles ? 5. How many times will a wheel, 16 feet 6 inches in cir- cumference) turn round in go- ing from Boston to Providence, it being 40 miles ? 7. If a man step 2 feet mches at once, how many steps will he take in walking 43 miles ? i . ; 9 ^^J^' <^ ^ ; ' ^-^ ' Cloth Measure. 4. In 30539520 inches, how many miles ? 6. If a wheel, 16 feet 6 inches in circumference, turn round 12800 times in going from Boston to Providence, what b the distance ? a A man walked 90816 steps, of 2 feet 6 inches each, in a day; how many mile^ did he walk? \ ir 119* Cloth Measure is a species oC linear measure, being used to measure cloth and other goods sold by the yard in length, without regard to the width. The denominations are ells, yards, quarters, and nails. 4 nails, (na.,) or 9 inches, make 1 quarter, marked qr. 4 qrs., or 36 inches, 1 yard, " yd. 3qrs. 1 ell Flemish," E.Fl. 5 qrs. 1 ell English, " E.E. 6 qrs. 1 ell French, " E.Fr. NoTs. — Eighths and sixteenths of a yard are now used instead of nails. SXAMPUES FOR PRACTICE. 1. In 673 yds. 1 qr. 1 na., how many nails? 3. In 296 E. E. 3 qrs., how many nails ? 5. In 151 E. E., how many yards? 7. In 29 pieces of cloth, each containing 36 E. FL, how many yards ? 2. In 9 173 nails, how many yards ? 4. In 5932 nails, how many E.E.? 6. In 188 yds. 3 qrs., how many E. E. ? 8. In 783 yds., how many E. FL ? Questions. — IT 1 10. What is cloth measure ? How used ? Why a sx^ecies of lini^ar measure ? Give the denominations, and the sign of each; the table. What is used instead of nails? How may yaitis be Tcdueed to £. £. ? to £. Fr. ? to £. Fl. ? How each of these to yards? 140 COMPOUND NUMBERS. iriu. 1 • 2 ' 3 3 feet = 1 yard. 1 II. Land or Square Measure. IT 11 1* Sqfiare Measure is used in measuring land, and other things wherein length and breadth are considered. 1 linear yard. c NoTE. — It takes 3 feet in length to make 1 \ linear yard. But it requires a square, 3 feet = 1 linear yard in length, and 3 feet =: 1 linear yard in breadth, to make 1 square yard. 3 feet in length and 1 foot in width, make 3 square feet, (3 squares in a row, TT 48.) 3 feet in length and 2 feet in width make 3X^ = 6 square feet, (2 rows of squares, TT 48.) 3 feet in length and 3 feet in width make 3X3^9 square feet, (3 rows of squares.) It is plain, also, that 1 square foot, that is, a square 12 inches in length and 12 inches in breadth, must contain 12 X 12 = 144 square inches, (12 rows, of 13 squares each.) The denominations of square measure are miles, acres, roods, rods or poles, yards, feet, and inches. TABLE. 144 square inches (sq. in.) make 1 square foot, marked sq. ft. • I ^^ 1 .1 eo • 9 sq. ft. = 1 sq. yd. 9 square feet 1 square yard, 30J sq. yds.=:5J X 5J> ^^ 1 J ^ ^^* ^^^» perch, 272| sq. ft. = 16 J X 16 J, ) ( or pole, 40 square rods 1 rood, 4 roods, or 160 square rods 1 acre, 640 acres 1 square mile, EXAMPIiES FOR PRACTICS. (i r (( ii (i sq. yd. sq. rd. P. R. A. M. 1. In 17 acres 3 roods 12 poles, how many square feet ? 3. Eeduce 64 square miles to square feet. 5. There is a town 6 miles square; how many square miles in that town? how many acres ? 2. In 776457 square feet, how many acres ? 4. In 1,784,217,600 square feet, how many square miles ? 6. Reduce 23040 acres to square miles. Questions. — If 111. What is square measure? What is a square ? Draw or describe a square inch ; a square yard ; a figure showing the square inches contained in a square foot. How many square inches in a row, and how many rows of square inches would it contain ? Multiply questions at pleasure. What are the denominations of square measure? Repeat the table. How does square measure differ from linear measure ? T 112, 113. COMPOUND NUMBERS. 141 7. How many square feet 8. In 5510528179200000 on tbe surface of ^e globe, square feet, how many square supposing it contain 197,663,- nules ? 000 square miles ? V f 13« The Surveyor's f or what is called GnnUr's Chain, is generally used in surveying land. It is 4 rods, or 66 feet in length, and consists of 100 links. TABIM FOR lilNKAR MSASURK. 7^^ inches make 1 link, marked L 25 links 1 rod, " rd. 4 rods, or 66 feet, 1 chain, " C. 80 chains 1 mile, " mi. 1. In 5 mi. 71 C, how 2. In 471 chains, how many chains ? many miles ? . 3. Reduce 2 mi. 15 C. 3 4. Reduce 17593 links to ids. 18 1. to links. miles. 5. In 75 C, how many 6. In 4950 feet, how many feet ? chains ? TABIJC FOR SaUARB MSASURE. 625 square links (sq.l.) .ake j l^l^X. \ """'' ""'l^'' 16 square poles 1 square chain, " sq. C. 10 square chains 1 acre, " A. OTE. — Land is generally estimated in square miles, acres, roods "^ uare poles or perches. Reduce 8 A. 2 sq. C. 7 8. In 824831 sq. 1., how P. 456 sq. 1. to square many acres ? links. 9. In 80 A., how many 10. In 8000000 sq. 1., how square chains ? how many many square chains ? In 800 square links ? sq. C., how many acres? III. Cubic or Solid Measure. IT lis. Cubic or Solid Measure is used in measuring things that have length, breadth, and thickness ; such as tim- ber, wood, earth, stone, &c. Questions. — If 1 12. What is generally used in surveying landl What is its length? Of how many links does it consist? Repeat the table for linear measure ; for square measure. How is land geneially estimated? '- - ■ m COUPODND NUU tll3. Hon 1. — It hu been shown (\111,) tliat 1 eqnue jrard contuiw S X 3 = S Bqu&re feet. . _ . A block 3 feet long, 3 feet KJy wide and 3 feet thick, is a *• cubic yard. The accompanying Igure represents soch a block. Were a portion 1 foot in thick- } nets cut off_from the top of this j block, the part cut off would be r 3 feet long, 3 feet wide, and 1 foot f thick, and would contain 3x3 X 1 ^ 9 cubic feet. The bottom part being 3 feet long, 3 feet wide, and 2 feet thick, would contain-3 X 3 X 2 = 18 cubic feet. But the entire block being 3 feet long, 3 feet wide, and 3 feet thick, contains 3 X 3 X 3 = 27 cubic feet. It is plain also, that a cubic foot, that is, a solid body, 12 inches longi 12 inches wide, and 13 inches thick, will contaia 12 X 12 X 12^ 172S cubic or solid inches. The denominations of cubic measure are cords, tons, yardst feet and inches. ■ 1728 cubic inches, (cu. in.) = 12 X 12 X 12, that is, 12 inches in length, 12 ill breadth, ^nd 12 ia thickness, _ 27 cubic feet, 3x3x3, SO feet of round timber, or 40 feet of hewn timber. 42 cubic feet 8 cord feet, or ) 128 cubic feet, ( make 1 cubic foot, parked cu. it 1 cubic yard, " ^Byd. 1 ton, " T. ( 1 ton of shipping, ) rn !1 cord foot, or I (^ ^ 1 foot of wood, i *^- "• 1 cord of wood, C. Qnestione. — If 113. fPhat is cubic measnre? What distinctions do yoa mabe between a line, a, surface and a, solid? What is a cube I a cubic inch ! a cubic foot ? a cubic jaii I For what is cubic or eolid measure used? What are its denomiaaiioas? Repeat the table. For what is (he cubic ion used, t What do jon understand by a ton of round timber ! "Vhat are the dimensions of a pile of wood coniaining 1 ccod t What is a cord foot? fll4 COMPOUND NUMBERa 143 NoTB 2. — A cubic ton is used for estimatkig the oaitage and trans- portation of timber. A ton of round timber is such a quantity (about fi(^feei) ae will make 40 feet when he^Krn square. Note 3. — A pile or load of wood 8 feet long, 4 feet wide, and 4 feet high, contains 1 coid. 8 X 4 X 4» 1^ txxUo feet. A cord foot is 1 foot in length of snoh a pile. 1. Eeduce 9 tons of round timber to cubic inches. 3. In 37 cord feet of wood, how many solid feet ? 5. Reduce 8 cords of wood to cord feet. 7. In 16 cords of wood, bow many cord feet? how many solid feet ? 9. In 26 C, 5 C. ft., 9 cu. ft., 1575 cu. in. of wood, how many cubic inches ? 2. In 777600 cubic inches, how many tons of round tim- ber? 4. In 592 solid feet of wood) how many cord feet ? 6. In 64 cord feet of wood, how many cords ? 8. 204S solid feet of wood, how many cord feet? how many cords ? 10. In 5684967 cubic inches, how many cords ? Measures of Capacity. I, Wine Measuee. IT 1 14L* Wine Measure is used in measuring all liquids except ale, beer, and milk. The denominations are tuns, pipes, hogsheads, tierces, bar- rels,gallons, quarts, pints, and gills. ^ TAB1.I:. 4 gill^ (gi.) make 1 pint, ^ pi^^Lf 1 quart. 4qut 31J gallons 42 gallons 63 gallons, or 2 barrels, 2 hogsheads 2 pipes, or 4 hogsheads. 1 gallon, 1 barrel, 1 tierce, 1 hogshead, 1 pipe, 1 tun, marked pt. qt. u (( 44 {{ 44 gal. bar. tier, hhd. P. T. NoTB. — The wine gallon contains S31 cubic inches. A hogshead 3f molasses, ^c , is no definite quantity, but is estimated by the gal- lon. C^nesti ons • — IF 11 4. For what is wine measure used ? What are Its denominations ? Repeat the table. How many cubic inches in a wine gallon ? How many gallons in % hogshead of molasses ? Ace 144 COMPOUND NUMBERS. IT 115, 116. 1. Reduce 12 pipes of wine 2. In 12096 pints of wine to pints. how many pipes ? 3. In 9 P. 1 bhd. 22 gals. 4. Reduce 39032 gills to 3 qts., how many gills? pipes. 5. In 25 tierces, how many 6. In 33600 gills, how gills ? many tierces ? n. Beeh Measure. IT 1 IS* Beer Measure is used in measuring beer, ale and milk. The denominations are hogsheads, barrels, gallons, quarts, and pints. TABUS. 2 pints (pts.) make 1 quart, marked qt. 4* quarts 1 gallon, " gal. 36 gallons 1 barrel, " bar. 54 gallons, or 1 J barrels, 1 hogshead, " hhd. Note. — The beer gallon contains 282 cubic inches. 1. Reduce 47 bar. 18 gal. 2. In 13680 pints of ale, of ale to pints. how iQiiny barrels ? 3. In 29 hhds. of beer, ~ 4. Reduce 12528 pints to how many pints ? hogsheads. in. Dry Measure. fl" 116« Dry Measure is used in measuring all kinds of grain, fruit, roots, (such as carrots and turnips,) salt, charcoal, &c. A The denominations are chaldrons, quarters, busheli^ecks, quarts, and pints. TABIifi* 2 pints (pts.) make 1 quart, marked ^ 8 quarts 1 peck, " pk. 4 pecks 1 bushel, " bu. 8 bushels 1 quarter, " qr. 36 bushels 1 chaldron, " ch. Note 1. — The dry gallon contains 268^ cubic inches. The Winchester bushel, which is adopted as our standard, contains 2150f cubic inches. It is 18^ inches in diameter, and 8 inches deep. The quarter of 8 bushels is an EngUsh measure. Questions. — 1[115« Wliat is the use of beer measure? What are its denominations ? Repeat the table. How many cubic inches in • beer gallon? f ^ T 117. COBIPOtJND NUMBERS. 145 Note 3. — The ImpeTial gallon, adopted in Great Britain in 19SCf for all liquids and dry sabetaoces, contains 277^(f(f cubic inches. 1, In 75 bushels of wheat, 2. In 4800 pints, how how many pints ? many bushels ? 3. Reduce 42 chaldrons of 4. In 6048 pecks, how coal to pecks. many chaldrons ? 5. In 273 qrs. 6 bu. 3 pks. 6. In 140223 pints, how 7 qts. 1 pt. of wheat, ho^ir many quarters ? mttny pints ? 1 hour. « h. 1 day, M d. 1 week, « w Time. V 117* Time is the measure of duration. The denominations are years, months, weeks, days, hour8« minutes, and seconds. TABLiB. 60 seconds (s.) make 1 minute, marked m. 60 minutes 24 hours 7 days 62 weeks 1 day 6 hours 48 min- L utes 48 seconds, or 365 days 5 ^ 1 year, " yx. f hours 48 minutes 48 seconds, Note 1. — As there is nearly | of a day more than 365 dsjB in a year, we add 1 day to February of certain years, thus giving them 366 days. K the excess was just J of a day, we would add 1 day to every 4th year, thus making the years average 365 days 6 hours, the odd day being added to every year exactly divisible by 4. But as the excess is not quite 6 hours, lacking about | of a day in 100 years, a year divisible by 100, though divisible by 4, has only 365 days, unless it be divisible by 400, when it has 366 days. Thus, 1844 and 1600 had 366 days each, but 1845 and 1700 had only 365 days each. A year of 366 days is called Bissextile, or Leap year. The calendar months, into which the year is divided, are from 28 to 31 days in length. Questions. — If 1 16. Eor what is dry measure used ? Wiat are its denominations ? Repeat the table. How many cubic inches in a dry gallon? Describe the Winchester bushel. What measure is the quar- ter? What is said of the Imperial gallon ? Which is the larger quan- tity, a quart of milk, or a quart of sidt ? — a quart of milk, or a quart of Tinegar ? — a quart of oats, or a quart of cider ? 146 COMPOUND NUMBER& iriia The number of days in each month may easily be remembered from the following lines : Thirty days hath September, April, June, and November ; All the rest have thirty-one, Save February, which alone . Hath twenty-eight ; and one day moie We add to it, one year in four. EXAMPLES FOR PRACTICE. 1. "How many seconds from * Jan. 1, 1790, till March 1, 1804, including the two days named, and making allowance for leap years ? 3. How many minutes from July 4th, M., to Sept. 29th, 6 o'clock, P. M. ? 6. At Boston, on the long- ^.est days, the sun rises at 23 min. past 4 o'clock, and sets 40 min. past 7; how many seconds in such a day ? 7. How many minutes from the commencement of the war between America and Eng- land, April 19th, 1775, to the settlement of a general peace, which took place Jan. 20th, 1783? Note 9. — The pupil will no- tice that the years 1776 and 1780 were leap years. 2. In 446947200 seconds how many weeks ? 4. On what month, day, and hour, will 125640 minutes past 12 o'clock, M., July 4th, expire ? 6. In 55p20 seconds, how many hours ? 8. In 4079520 how many years ? minutes, Circular Measure, or Motion. IT 118« Circular Measure is used in computing latitude and longitude ; also in measuring the motions of the earth, and other planets round the sun. Questions* — ^ 117* Of what is time the measure? What are the denominations ? Repeat the table. Why is 1 day added to Feb. of certain years? Why is it not added to every 4th year? What years have 365 days, and what 366 days ? What is a year of 366 days tailed f Name the ealendar months, and the number of days in each. T119, 12a COMPOUND NUMBERa 147 The denomintitions axe ciiclesi 8ign8» degrees9 minntes, and seconds. 60 seconds {") make 1 minnte, marked \ 60 minutes 1 degree, * ** •. 30 degrees 1 sign, ^* S. 12 signs, or 360 degrees, 1 circle. 1. Reduce 9st 12? 25' to 2. In 1020300'% howjnany seconds. degrees ? 3. In 3 signs, how many 4. In 5400% how manjf minutes ? signs ? iril9« Miscellaneoiui Table. E 20 units make 1 score. ^ 100 lbs. of raisins make 1 cask. 5 score 1 hundred. 100 lbs. of fish 1 quintal 12 units 1 dozen. 100 lbs. 1 hundred 12 doz. s= 144 1 gross. 18 inches 1 cubit 12 gross ss 144 doz. 1 great gross. 22 inches, nearly, 1 sacred cubit 200 lbs. of beef, j . , , 1 gallon of tram oil 7J lbs. port:, or fi^ J oairei. ^ gallon of molasses 11 lbs. 196 lbs. of flour 1 barrel. 24 sheets of paper 1 quire 8 bushels of salt 1 hogshead. 20 quires 1 ream. 280 lbs. of salt at I .2 reams 1 bundle the salt works > 1 barrel. 5 bundles 1 bale. m N. Y. ) A sheet folded in two leaves, or 4 pages, is called a folio A sheet folded in four leaves, or 8 pages, a quarto, or 4to. A sheet folded in eigh|leaves, or 16 pages, an octavo, or 8vo. A sheet folded in twelve leaves, or 24 pages, a duodecimo, or 12mo. A sheet folded in 18 leaves, or 36 pages, an 18mo. A sheet folded in 24 leaves, or 48 pages, a 24mo. 5 points make 1 line, ) used in measuring the length of the rods of 12 lines 1 inch, ) clock pendulums. 4 inches 1 hand, used in measuring the hight of horses. 6 feet 1 fathom, used in measuring depths at sea. Reduction of Fractional Compound Numbers. IT 190* There are four particular cases in the reduction of fractional compound numbers. 1st, To reduce a fraction of a higher denomination to one of a lower, 2d, To reduce a fraction of a lower denomination to one of a higher. 3d, To Questions. — If 118. What are the uses of circular measure Wbiu ar« the denominations? Bepeat the table. it L„^^ 148 COMPODND H«,j:i-'1.'^. f 180. reiuee afrttotion of a high dewmmatim to miegers of lower denominations. 4th, To reduce integers of Unoer denomifna* tions to a fraction of a higher. We will consider them in their order. I. To reduce a fraction of ^ a higher denominaHon to- one of a lower. 1. Reduce ^ of a pound to the fraction of a penny. Solution. — We feust reduce 2-J^ of a pound to the fraction of a shilling by multiplying it by 20, since 20 shillings make J^l. This done by ^ T6, gt^es ^ of a shilling, which multiplied by the composite number 12, (^ 76, note 1,) is reduced to the fraction f- of a penny. Hence, Multiply as in the reduction of whole ^numbers, according to the rules for the multiplica- tion of fractions. n. To reduce a fraction of a lower denomiTUEtum to one of a higher. 2. B^dwieJ- of a peooy tP the fraction of a pound. ' Solution. — We must reduce •^ of a penny to the fraction of a shiUing, by dividing it by the composite number 12, since 12 pepce make 1 shilling. This done by T 81» note 2, gives -^ of a shilling, which divided by 20, (multiplying the denominator,) is reduced to the fraction ^^ of a pound. Hence, RUIiE. Divide as in llie reduction of whole numbers, according to the rules for the division of fractions. BXAMPIiES FOR PRACTICE. 3. Reduce yf ^j- of a pound of gold to the fraction of a grain. 5. Reduce ttVv of a hogs- head of milk to ihe fraction of a pint. 7. Reduce A of a hogs- head of ale to tne fraction of a barrel. 9. Reduce ^'j^^h ^^ ^ ^^ of oil to the fraction of a gill. 4. Reduce ^^ of a grain of gold to4|^e fraction of an ounce. 6. Reduce ^§f of a pint of milk to the fraction of a hogs- head. 8. Reduce ^ of a barrel of ale to the fraction of a hogshead. 10. Reduce if ill of a gill of oil to the fraction of a tun» Questions* — If 120* How many cases in the reduction of fractional compound numbers ? Give the first. How are integers reduced from a higher denomination to a lower ? — from a lower denomination to a higher? How are fractions reduced from a higher denomijiation to a lower ? Give the example and its explanation. Rule. How are they reduced ' firom a lower denomination to a higher ? Give Ex. 2 and t» aolution. Rule. ' I 1 Tidi. OOHPODND NUMBBRa 141 11. Redace of a square mJe to the frac- tion of a square inch. 13. Reduce t^Vv ^ ^ bushel to the fraction of a pint. 15. Reduce ttiWv ^^ & week to the fraction of an hour. • 17. A cucumber grew to ^e length of 79^ of a mile ; what part is tk^t of a foot ? 19. Reduce f of | of a pound to the fraction of a shillinfif. 21/Reduce i of ^8> of 3 pounds to the fraction of a penny? ITISI. in. To reduce a fraction of a high denomina- Hon t^^ integers of lower de- nominations, 1 . How many shillings and pence in § of a pound ? Solution. — Multiplying $ of a pound by 20, it is leduced to the fractioii of a shilling, ^. But as ^ of a shilling is an improper fraction, (^ 65,) it contains sever- al shillinfs. The whole shillings ^e find, dividing the nuixMrator by the denominator, to be 13, and a fraction, |, of a shilling remains, and this redueed to the fraction of a penny, is -^ of a penny = 4d. Hence ISs. 4d. 19 the Ans, 13* 12. Redace tVWA WW W fli of a square inch to the frac- tion of a square mile. 14. Reduce A of a pint to the fraction of a bushel. 16. Reduce ^ of an hoar to the fraction of a week. 18. A cucumber grew to die length of 1 foot 4 inches sBB^s=s| of afoot; what part is that of a mile ? 20. }f of a shiUing is I of what fraction of a pound ? 22. ^ffL of a penny is | of what fraction of 3 pounds? J|*|ft of a penny is ^ of what part of 3 pounds ? -^f^ of a penny is ^ of yV ^^ ^^ many pounds ? IV. To reduce integers of lower denominations to afrac" tion of a higher denomination. 2. What part of a pound is 13s. 4d. ? Solution. — In a whole pound there are 240 pence, and we wish to find what part of this number of pence is contained in 13s. 4d. 13s. 4d. reduced to pence, is 160 pence. Hence 13s. 4d. is 160 out of 240 pence contained in a whole pound, or j^ = ] of a pound, Ans, Note 1 . —The nujnerator and denom- inator of a fraction most be of the same denomination, sinee the fj^rmer is a dividend, and the latter a divisor, both of which must be of one denomination, (1133.) Hence, if there is a frac- tional part to the integer of the lowest denomination, for example, were it required to reduce 4d. 3lqi8. to the fraction of a shilling, we should have to reduce U. to Sdsof a fiffthing form 160 COMPOUND NUMBERS. f 121. Herce, Ruur. Reduce the given fraction to the next lowpf denomina- tion , and, if it r then an im- proier fraction, reduce it to a whole or mixed number,— the integer is the number of this denomination. If a frac- tion remains, reduce it, as before, to the next lower de- nomination. So proceed, if a 'fraction continues to remain, to the lawest denomination. denominator, and 4d. 3lqrs. to Sds of a farthing for a numerator. The for- mer will then show the number of 3ds of a farthing in Is., the latter how many of them are contained in id. Siqrs. Hence, Reduce the. given sum to the lowest denomination con- tained in it for a numerator, and a unit of the required higher denomination to the same denomin'^.cion for the denominator. SXAMPUBS FOR PRACTICB. 3. Iteduce f of 1 day to liourh and minutes. OPERATION. Numer, 3 24 4. Reduce 14h. 24min. to the fraction of a day. OPERATION. 14A. 24 min, 1 da. 60 24 Ikwm. 5 ) 72 ( 14ft. 24w. Am. 864 NuTner. 24 5 60 22 20 2 60 1440 Denom, ff^^=^iday. Am. 120 10 20 20 Questions. — ^ 121* What is case III.? Give the solution; the lie. Case IV. ; the solution j the rale. Why must the numerator and enominator of a fraction be of the same denomination ? What follows, 4sn, in case of a fraotional part? I 1123. COMPOUND NUMBERa Ifil 6. What is the value of f 6. Reduce 7 oz. 4 pwt to of a pound, Troy ? - the fraction of a pound, Troy. 7. What is the value of f 8. Reduce 8 oz. 14f dr. to of a pound, avoirdupois ? the fraction of a pound, avoir- dupois. 9. Reduce f of a mile to 10. Reduce 4 fur. 125 yds. its proper quantity. 2 ft. 1 in. 21 har. to the frac- tion of a mile. 11. I of a week is how 12. 5 d. 14 h. 24 m. is many days, hours, and min- what fraction of a week ? Qtes? 13. Reduce -/g- of an acre 14. Reduce 1 rood 30 poles to its proper quantity. to the fraction of an acre. 15. W hat is the value of 16. Reduce 2 ft. 8 in. 1^ h. ^ of a yard ? to the fraction of a yard. Note. — Let the pnpil be lequiied to levene and piOTO the follow ing examples : 17. Reduce 3 roods 17} poles to the fraction of an acre. 18. A man bought 27 gal. 3 qte. 1 pt. of molasses ; what part is that of a hogshead ? h^ 'I ' 19. A man purchased ^ otTcvrt. of sugar; how much WLg&x did he purchase ? J - 2 • ? / ,^ . 20. 13 h. 42 m. 51f s. m what part or fraction of a day ? /' Reduction of Decimal Compound Numbers. ir 1S3« I. To reduce the II. To reduce integers of decimal of a higher denomiruu' lower denominations to adeci" tion to integers of htoer de* mal of a higher denomination, nominations. 1. Reduce *375£ to inte- 2. Reduce 7s. 6d. to the gers of lower deiominations. decimal of a pound. OPERATION. •375£. OPERATION. 20 12 6*0 7*500^. 20 12 7*500 6*000£?. Ans. 7s. 6d. *375 of a pound, Am Solution. — Multiplying *375 Solution. — We divide 6d. by of a pound by 20, it is reduced to 13, to reduoo it to shillings, bat 1^ COMPOUND mJHOBBS. V109. 7*5008., observing the cnrdinary rule for pointing of decimals in the product, that is, 7 shillings, and *500 of a shilling. This decimal multiplied by 19 becomes 6'OOOd., that is, 6d. and no deci- mal. Hence *375 of a pound =s 7s. 6d. Ana. Hence, RUUD. Multiply the given decimal by the number which will re- duce it to the next lower de- nomination, pointing off deci- mals according to the ordinary rule; reduce thi^ decimal to the next lower denomii^ation, pointing off as before. So continue to do through all l^e denominations ; the several integers will be those quired. re- as it wiU not make a whole shil- ling, we annex a cipher to reduce it to lOths, (1 91 ;) then 12 in 60 tenths, 5 tenths of a shilling ; annexing this to 7 phillings, we have 75 tenths of a shilling to re- duce to the dedimal of a pound, and dividing by 20, annexing ciphers to leduce the remainder to hundredths and thousandths, we have *375 of a pound. Hence, RULOU Divide the lowest denomi- nation, annexing ciphers as may be necessarj^, by the number which will reduce it to the next higher, and annex*- ing the quotient to the num- ber of this higher denomina- tion, divide as before. So co|}tinue to do till the whole is brought to the required decimal. EXAMPIiES FOR PRACTICIU 3. Reduce *213 of a "long ton " to integers of lower de- nominations. 5. Reduce *6 of a lb. of emetic tartar to integers of lower denominations. 7. In *76754 of a square mile, how many integers of lower denominations ? 9. Reduce *3958 of a bar- rel of wine to integers of low- er denominations. 4. Reduce 4 cwt. 1 qr. 1 lb. 1 oz. 14*72 drs. to the deci- mal of a " long ton." 6. Reduce 7 g. 1 5. 19. 16 grs. to the decimal fraction of 1 lb. 8. What decimal of a square mile is 491 acres 36 square rods 26 square feet 19*584 square inches ? 10. Reduce 12 gal. 1 qt. 1 pt. 2*9664 gills of wine to the decimal of a barrel. Questions. — If 123« How many cases in the reduction of deci- mal compound numbers? What is case I.? Give the example and its solution. Give the rule. What is case XL? — the solution of the ex- ample ? ^ the rule? f U&. COMPOUND NUIOBEUl 1A3 11. How many integers of 12. What decimal of a cord lower denominations is '73 of is 5C. ft. 13 cu. ft. 760'32 cu. a cord ? inches ? 13. In *648 of a quarter of 14. In 5 bu. 5 qts. 1*776 wheat, how many integers of pts. of wheat, what fraction a less denomination ? of a quarter ? 15. Reduce '125 lbs. Trojr 16. Reduce 1 oz. 10 pwt to integers of lower denomi- to the fraction of a pound, nations. 17. What is the value oi 18. Reduce 38 gals. 3'52 '72 hhd. of beer ? qts. of beer, to the decimal of a hhd. 19. What is die value of 20. Reduce 1 qr. 2 na. to *375 of a yard ? the decimal of a yard. 21. What is the value of 22. Reduce 17 h. 6m. 43^ '713 of a day ? sec. to the decimal of a day. Let the pupil be required to rererse and prove the following ex* amples: 23. Reduce 4 poles to the decimal of an acre. 24. What is the value of *7 of a lb. of silver ? s 25. Reduce 18 hours 15 m. 50*4 sec. to the decimal of a day. 26. Reduce 11 mi. 6 fur. 2 rods 3 yds. 2 ft. to the decimal of a degree on the equatorial circumference of the earth. f ISS. Review of Reduction of Compound Numbers. Qoeations. — What are compound numbers? What is meant bj the word denomination? What is reduction of compound numbers? What are the kinds, and how performed? Changing ells Eng. to yards is reduction — what kind ? What is the use and what the denomina- tions of Troy weight ? Avoirdupois weight ? Which is larger, 1 oz. ^ Troy, or 1 oz..Avoirdupois?— 1 lb. Troy or 1 lb. Avoirdupois? What ^ distinction do you make between the long and the short torij and where are the two used? What distinctions do you nMike between linear, j square, and cubic measure? What are the denominations in linear measure? — in square measure? — in cubic measure? How dc you j multiply by i ? When the divisor contains a fraction, hfsm do you pro- j cecd ? How is the superficial contents of a square figure found ? How I is the solid contents of any body found in cubic measure ? How many j Solid or cubic feet of wood make a cord? What is understood by a cord foot? ilow many such feet make a cord? How many rods in length IS (hunter's chain? Of how many links ^oes it consist? How many links make a rod? How many rods in a mile? How many sqnavt rods in «n acre f 1 154 COMPOUND NUMBERa H 123. How many pounds make 1 cwt. ? For what is ciitnlar measure nsed f Into Qow many parts is a smaU ciide divided ? — a large circle? — called EXBRCISBS. 1. In £46 4s. sterling, how many dollars ? (Consult IT 104, note 3.) Am. $223^608. 2. How many rings, each weighing 6 pwt 7 grs., may be made of 3 lb. 5 oz. 16 pwt. 2 grs. of gold ? Am, 158. 3* Suppose West Boston bridge to be 212 rods in length, how many times will a chaise wheel, 18 feet 6 inches in cir- cumference, loirn round in passing over it ? Ans. 189^ times. 4. In 10 lb. of silver, how many spoons, each weighing 6 oz. 10 pwt. ? , Atis, 21-j»t spoons. 5. How many shingles, each covering a space of 4 inches one way, and 6 inches the other, would it take to cover 1 square foot ? How many to eover a roof 40 feet long, and 24 feet wide? (See IT 48.) Arts, to the last, 5760 shingles. 6. How many cords of wood in a pile 26 feet long, 4 feet wide, and 6 feet high ? Ans. 4 cords, and 7 cord feet. 7. There is a room 18 feet long, 16 feet i^de, and 8 feet high; how many rolls of paper, 2 feet wide and 11 yards long, will it take to cover the walls ? Ans, 8^. 8. How many cord feet in a load of wood 6^ feet long, 2 feet wide, and 5 feet high ? Ans, 4^ cord feet. 9. If a ship sail 7 miles an hour, how far will she sail in 3 w. 4 d. 16 h. ? * 10. A merchant sold 12 hhds. of brandy, at $2'75 a gal- j Ion ; what did he receive for each hogshead, and to how much did the whole amount ? 11. A goldsmith sold a tankard for £10 8s. at the rate of 5s. 4d. per ounce ? how much did it weigh ? A?is. 3 lbs. 3 oz. 12. An ingot of gold weighs 2 lb. 8 oz. 16 pwt. ; how much is it worth jit 3d. per pwt. ? \ • 13. If a cow give, on an average, 9 qts. of milk each day, how much will she give in a year, or 365 days ? Am, 15 hhd. 11 gal. 1 qt. 14. Reduce 14445 ells Flemish to ells English. 15. There is a house, the roof of which is 441 feet in length, and 20 feet in width, on each of the two sides ; if 3 shingles in width cover one foot in length, how many shin- gles <viil it take to lay one course ? If 3 courses make one / T 123. COMPOUND NUHBEBa 155 foot, how many courses will there he on one side of the roof? How many shmgles will it take to cover one side ? to cover both sides ? Ans, 16020 shingles. 16. How many steps, of 30 inches each, must a man take in traveling 54^ miles ? 17. How many seconds of time would a person redeem in 40 years, by rising each morning ^ l^our eanier than he now does? , ' ' ' V ^ 18. If a man lay up 4 shillings each day, Sundays ex- cepted, how many dollars would he lay up in 45 years ? . 19. If 9 candles are made from 1 pound of tallow, how many dozen can be made from 24 pounds ? 20. If one pound of wool make 60 knots of yam, how many skeins, of ten knots each, may be spun from 4 pounds 6 ounces of wool ? Ans. 26| skeins. 21. How many hours from the commencement of the com- mon Christian era till Dec* 10, 1847, 12 o'clock, noon, allow- ance being made for leap years ? How many weeks ? ^ . ( 16189932 hours. ^Ans. I 96369^ ^^eks. 22. What part of a pwt. is y^^y of a pound Troy ? Am, ^ pwt • 23. What fraction of a pound is | of a farthing ? Am, ^£t2Vt>« 24. What fraction of an ell English is 4 qrs.*lj na. ? Am. i E. E. 25. What fraction of a yard is 2 qrs. 2§ na. ? Am, § yd. 26. What fraction of a day is 16 h. 36 m. 55^ s. ? Ans, ^ d. 27. What fraction of a mile is 6 fur. §6 r. 11 ft. ? Atis, ^ mi. 28. What is the value of -j^ of a " long ton ? '* Am. 4 cwt. 2 qrs. 12 lb^ 14 oz. 12^ drs. 29. What decimal of a- day is 55 m. 37 sec. ? Am, *03862+d. 30. What decimal of a pound Troy is 10 oz^Q pwt. 9 gr> ' A?Jr'889Q625. 31. What is the value of *397 of a yaifl ? Ans, 1 qr. 2 na. + lb. ♦Consultir 117, Note 1. 156 COMPOUNP NUMBiPP. T IM Addition of Compound Numbers. IT 124. 1. A boy bought a knife for 1 shilling 9 pence, and a comb for 1 shilling 6 pence ; how much did he give for both ? Am. 3s. 3d. 2. A grocer sold at one time 2 qts. of molasses, at another time 3 qts., at another 1 qt., at another 3 qts., and at another 2 qts. ; how many gallons did he sell ? Solution. — 3 qts. + 2 qts. + 1 qt. -|- 3 qts. -f- 2 qts. = 11 qts., and 11 qts. =s 2 gal. 3 qts. Ans, 2 gal. 3 qts. 3. A boy had 30 rods to walk ; he walked the first 10 rods in 30 seconds, the next 10 rods in 45 sec, and the last 10 rods in 20 sec. ; how many minutes was he in walking the 30 rods ? Ans. 1 min. 35 sec. 4. What is the amount of 1 yd. 2 ft. 6 in + 2 yds. 1 ft. 8 m. ? A7i$. 4 yds. 1 ft. 2 in. 5. A man has two bottles which he wishes to fill with wine ; one will contain 2 gal. 3 qts. 1 pt., and the other 3 qts. ; how much wine can he put in them ? Ans. 3 gal. 2 qts. 1 pt. The uniting together in one sum of several compound numbers is called Compound Addition. 6. A man bought a horse for £15 14s. 6d., a pair of oxen for £20 2s. 8d., and a cow for £5 6s. 4d. ; what did he pay for all ? Solution. — As the numbers are large, we write them down, placing those of the same denomination under each other, syid, begin- ning with those of the least value, add up each kind separately ; thus: — Then, adding up the pence, we find the OPERATION. amount to be 18, which we divide by 12 to re- ic -iA a ^^°® *^ shillings ; the remainder, which is 15 14 6 pence, we write* under the column of pence, i H 20 2 8 and add the 1 shilUng to the column of shil- "1 ) 5 ll^ 4 lings. Adding up the shillings, we reduce i [ — r^ — - them t© pounds; setting the remainder, 3 shil- \ \ Ans, 41 3 6 lings, under the column of shillings, we carry 4 I the one pound to the column of pounds, which '.,' \\ we add up, setting down the whole amount, as we do the amount of ;i |f the last column in simple addition. ^ I Note. — The operations in compound numbers difier firom those of U h simple numbers but in one particular. In simple numbers we aie ul f 194. QOIflffOUNb NITMBiatS. 1S7 gofeamad hf the law that %uiies inoioaae in a tenfold propoiCi<m from right to left, {% 9.) Compound numbera having no regular system of units, we are governed by the relations between the different de> nominations in which the several quantities are expressed. The above process is sufficient to establish the following RUUB Far Addition of Compound Numbers, I. Write the numbers so that those of the same denoniina- tion may stand under each other. n. Add together the numbers in the column of the lowest denomination, and carry for that number which it takes of the same to make one of the next higher denomination. Proc^d in this manner with all the denominations, till you come to the last, whose amount is written as in simple num- bers. Proof. -!- The same as in addition of simple numbers. l^lXAMPLi^ FOR PRACTICE 7. 8. Y. w. d. h. m. s. T. CUJt, or, lb. oz. dr. 67 7 6 23 55 11 14 U 1 16 5 10 84 8 16 42 18 26 2 11 9 15 32 24 5 5 18 5 7 18 25 11 9 9. Bought a silver tankard, weighing 2 lb. 3 oz., a silver cup, weighing 3 oz. 10 pwt., and a silver thimble, weighing 2 pv^rt. 13 grs. ; what was the weight of the whole ? Ans, 2 lb. 6 oz. 12 pwt. 13 grs. 10. A ship landed at N. York the following invoice of English goods, viz., 78 tons 3 cwt. 2 qrs. 26 lbs. of cotton ^ goods, 135 tons 15 cwt. 1 qr. 9 lbs. of iron, ^ tons 12 cwt. ^ 2 qrs. 20 lbs. of woollen goods, 225 tons 9 cwt. 17 lbs. of coal, * l and 106 tons 1 qr. of earthen ware ; what was the whole h amount? Ans, 636 tons 1 cwt. 16 lbs. ■' -■■ ■ ' " " ' ' — ■ I ■■■■■■ ■ ■». I I II ■■ I. ■ ■ , , . I ■ I ■ ..^ M P Questions, — 1[ 124. What is compound addition ? How are lower denominations reduced to higher ? (^ 105.) Repeat the rule for Qom- pound addition. How do you carry from farthings to pence?- -from pence to shillings? — from shillings to pounds? How do operations in j compound numbers differ from those in simple numbers? How do you carry through the several denominations of avoirdnpois weight? — long measure? 6^., te. * 14 4 . i t1 IQB COMPOUND NUMBERS. T 124. Note. — It will be recollected, (If 106,)* that wljat is called the '* long ton " is used in invoices of English goods, and of coal from Pennsylvania. 11. A boat took in freight as follows : at one place, 9576 lbs. of butter; at another, 11 tons of pork; at a third, 7 T. 18 cwt. 27 lbs. of coal ; what was the entire freight in " short tons ? " Am, €4 tons 1299 lbs. 12. A merchant bought 3 pieces of linen, measuring a& follows : 41 E. Fl. 1 qr. 2 na., 18 E. Fl. 2 qr. 3 na., 57 E. Fl. 1 na. ; how many Flemish ells in the whole ? Ans. 117E.F1. 1 qr. 2 na. 13. A draper bought 3 pieces of English broadcloth meas- uring as follows, viz., 75 E. E. 4 qr. 2 na., 31 E. E. 1 qr., 28 E. E. 1 na. ; how many English ells in the whole ? Ans. 135 E. E. 3 na. 14. There are four pieces of cloth, which measure as fol- lows, viz., 36 yds. 2 qrs. 1 na., 18 yds. 1 qr. 2 na., 46 yds. 3 qrs. 3 na., 12 yds. qr. 2 na. ; how many yards in the whole ? Ans. 114 yards. 15. A man travelled as follows, viz., the 1st day, 35 mi. 7 fur. 38 rd. ; 2d day, 4 mi. 2 rd. ; 3d day, 37 mi. 3 fur. 19 rd. ; 4th day, 44 mi. ; what was the length of his journey ? Ans. 121 mi. 3 fur. 19 rd. 16. Bought of Williams and Brother, London, 1 copy of Shakspeare for £7 14s. 6d., 1 copy of Arnold's Works for ^9 10s. 9d., 1 copy of the Edinburgh Encyclopedia for £27 6s., 1 quarto Bible for £8 6d., 1 copy of Johnson's Works for £15 2s. ; what did the whole cost ? Ans. £77 13s. 9d. 17. Bought at Liverpool 1 bale of cotton goods for £9 10s. 3d., 1 box of jewelry for £227 4s., 1 gross of buttons for £6 9s. 8d. ; what did I pay for the whole ? Ans. £243 3s. lid. 18. There are 3 fields, which measure as folio ws> viz., 17 A. 3 R. 16 P* 28 A. 5 R. 18 P., 11 A. 25 P. ; how much land in the three fields ? Ans. 58 A. 1 R. 19 P. ' 19. A raft of hewn timber consisted of 3 cribfe ; the 1st crib contained 29 T. 36 cu. ft. 1229 cu. in. ; the 2d, 12 T. 19 cu. ft. 64 cu. in. ; the 3d, 8 T. 11 cu. ft. 917 cu. in. ; how- much timber did the raft contain ? Ans. 50 T^ 27 cu. ft. 482 cu. in. •20. A man removed 79 cu. yds. 22 cu. ft. of earth in dig- ging a cellar, 9 cu. yds. 26 cu. ft. in digging a drain, and 22 cu. yds. 17 cu. ft. in digging a cistern ; how much earth did he remove? *An9^ 112 cu. yds. 11 cu. ft f 124. COMPOUND NUMBERS. 159 ■ 21. In one pile of wood are 37 cords 119 cu. ft. 76 cu. in. ; in another, 9 cords 104 cu. ft. ; in a 3d, 48 cords 7 cu. ft. 127 cu. in. ; in a 4th, ,61 cords 139 cu. in. ; how much wood in the four piles? Ans, 156 C. 102 h. 342 in. 22. A vintner sold in one week 51 hhd. 53 gaL 1 qt. 1 pt. of wine ; in another week, 27 hhd. 39 gal. 3 qts. ; and in another week, 9 hhd. 13 gal. 3 qts. ; how much did he sell in the three weeks ? ' Ans, 88 hhd. 43 gal. 3 qts. 1 pt. 23. . A milk-man sold in one week, 70 gal. 3 qts. of milk ; in another week, 67 gal. 1 qt. ; how many hogsheads did he sell ? Ans, 2 hhds. 30 gal. 24. A farmer sowed 36 hush. 2 pks. 5 qts. 1 pt. of wheat, and 19 bush. 3 pks. 7 qts. of barley ; how many bushels did he sow ? Ans. 66 bush. 2 pks. 4 qts. 1 pt. 25. A printer used in one week 6 bales, 7 reams, 9 quires and 9 sheets of paper, and in another week 14 bales, 9 reams, 19 quires and 15 sheets ; how much paper did he use in the two weeks ? Atis, 21 bales 7 reams 9 quires. 26. A ship sailed in one week as follows, viz., on Mon- day, 2^ 8' 45" south, P 51' east ; on Tuesday, 2*» 36' south, 2*> 1' 15" east ; on Wednesday, 4** 52" south, P east ; on Thursday, 1* 48' 52" south, 3* 16' 22" east ; on Friday, P 19' south, 48' 29" east; and on Saturday, 59' 30" south, 3* 52' 11" east; what was her distance south and east from the place of starting? . { South 13** 52' 59". ^^- I East 120 49/ 17// 27. A man plastered a church of the following dimensions, viz., the end walls contained 116 sq. yds. 7 sq. ft. 96 sq. in. each; the side walls, 178 sq. yds. 138 sq. in. each, and the ceiling 439 sq. yds. 6 sq. ft. 78 sq. in. ; what was the whole amount of plastering in the church ? Ans, 1029 sq. yds. 5 sq. ft. 114 sq. in. 28. What is the amount of 40 weeks 3 d. 1 h. 5 m. + 16 w. 6 d 4 m. -|- 27 w. 5 d. 2 h. ? Ans, 85 wk. 3 h. 9 m. 29. A miller sold flour as follows, viz., 4 bar. 176 lbs. 8 oz., 18 bar. 40J lbs., 1 bar. 104 lbs. 7 oz., 181| bar. ; how J' much did he sell in all ? Ans, 206 bar. 76 lbs. 7 oz. ;^ 30. Five bags of wheat weighed as follows, viz., 2 J bush., ; 2 bush. 21 lb. 7 oz., IJ bush. 18 lbs., 2 bush. 50 lbs., 1 bush. 58j lbs. ; what was their entire weight, calling 60 lbs. a . t bushel ? Ans, 1 1 bush. 13 lbs. 3 oz. ^ ? Note. — If the four following exampl^p^ correctfy wrought, the ; i results will be the same as those here given.' f > 160 COMPOUND NUMBfiBS. f 12$, 12t. 31. What is ditf sum of 35 bar. 27 gal. 3 qts. -f 19 bar. 5 gal. 1 qt. -}- 7 bar. 13 gal. 3 qts. ? Ans, 62 bar. 15 gal. 1 qt. , 32. What is the sum of 12 rd. 9 ft. 4 In. + 15 rd. 7 ft. 8 in. + 6 rd. 4 ft. 5 in. ? Ans. 34 rd. 4 ft. 11 in. 33. What is the sum of the following distances on the equatorial circumference of the earth, viz., 59 deg. 46 mi. 6 fur. 39 rds. 15 ft. 10 in. ; 216 deg. 39 mi. 7 fur. 39 rds. 4 ft. 7 in. ; 78 deg. 53 mi. 7 fur. 38 rds. 9 ft. 8 in. ? Afu. 355 deg. 2 mi. 4 fur. 11 rds. 2 ft. 7 in. 34. What is the sum of 2 A. 75 P. 248 sq. ft. 72 sq. in. -f 3 A. 120 P. 177 sq. ft. 85 sq. in. + 15 A. 17 P. 84 sq. ft. 80 sq. in. ? Ans, 21 A. 53 P. 238 sq. ft. 57 sq. in. 35. A hardware merchant sold several bills of screws, as follows: 25 great gross 9 gross 7 doz. 11 screws; 15 great gross 7 gross 8 doz. ; 40 great gross 4 doz. ; what was the whole amount sold ? Ans, 81 gr. gr. 5 gr. 7 doz. 11 screws. Addition of Fractional Compound Numbers. V ISti. 1. To ^ of an hour, add |^ of a minute. Solution. — First reduce each fraction to its proper quantity ; see ^ 121. I of an hour =» 52 min. 30 sec. ; | of a minute =s 52^1 seo. ; and 52 min. 30 sec. 4~ ^ sec. == 53 min. 22^ sec. Ans. 53 min. 22^1 see. Note. — It may sometimes be more convenient to reduce the frac- tions to the same denomination, add them together, and reduce their fractional sum to its proper quantity. 2. To I of a pound, add f of a shilling. Ans, 18s. 3d. 3. To ^ of a gallon, add J of a pint. Ans. 3 qts. 1-j*^ pts. 4. To ^ lb. Troy, add -^ of an oz. Ans. 6 oz. 11 pwt. 16 gr. 5. To fofa mile, add 47t^ rods. jItw. 239 rds. 4 ft. 6 in. 6. To S of 20 J yds., add 4 of 9^ yds. Am. 15 yds. 1 qr. 2^ na. Subtraction of Compound Numbers. IT 196* 1. From a piece of tape, containing 9 yds. 3 qrs., sold 4 yds. 1 q^ how much remained ? 2. A woman haviQg-6 Ibe. of butter, sold 3 lbs. 10 oz. * bow much had ^e left ? T 126. COKPOtJffD NUMBBMl 161 Solution. — Skioe ikme ue bo os. in the tot nmnber or raimiH eod, we take from 6 lbs. 1 lb. sb 16 oz. ; 16 oz. — 10 oz. = 6 oz., and 61b.— 3 Ib8. = 21b8. Ans. 2 lb». 6 oz. 3. How much is 1 ft. — (less ) 6 in. ? 1 ft. —8 in. ? 6 ft. Sin. — 1ft. 6 in.? 7 ft. 8 in. — 4 ft. 2 in.? 7 ft. 8 in.— 5 ft. 10 in. ? 4 How much is 4 weeks 3d. — 3 w. 4 d. ? 3 w. 1 d. — 2w. 5d.? Finding the difiereaee beenreen^Ciro oomponiid nomben, is sailed Compound Subtraction. 5. From 9 days 15 h. 30 m., take 4 d. 9 h. 40 min. oPBHATioN. Solution. — As the quantities are d. h, m, lar^e, it will be more convenient to Mtmeend, 9 15 30 wnte them down, the lees under the Subtrahend^ 4 9 40 greater, minutes under minutes, hours under hours, &c., since we must sub- Ans. 5 5 50 tract those of the same denomination from each other. We cannot take 40 m. from 30 m., but we may, as in simple numbers, borrow from the 15 h. in the minuend, 1 hour :» 60 minutes^ which added to 30 m. in the minuend, makes 90 m., and 40 m. firom 00 m. leaves 50 m., which we set down. Proceeding to the hours, having borrowed 1 from the 15 h. in the minuend, we must make this number 1 less, calling it 14 h., and say, 9 (hours) from 14 (hours) leaves 5, (hours,) which we set down. Lastly, proceeding to the days, 4 d. from 9 d. leaves 5 d., which we set down, and the work is done. 6. From £27, take £15 12s. 6d. OPERATION. Solution. — We have no pence from which to , take the 6d., but we must go to the pounds, and boi^ 27£. s, a, row JCl = 20s., and from the SOs. borrow Is. =* 12d. 15 12 6 Then 6d. from 12d. leave 6d. ; 12s. from 19s. (which 11 7 A remain of the £l) leave 7s. ; and J^15 from £26 ^^ ' ^ remaining, leave £IL Ans, J^ll 7s. 6d. The process in the foregoing examples may be presented m form of a . ' RUUS For Subtracting Compound Numbers. I. Write the less quantity under the greater, placing simi lar denominations under each other. II. Beginning with the least denomination, take the lowei 14* 164 COMPOUND mmsESBB. f las. 1. A note, beafing* date Dec. 86di, 1846, wa^ ]^aiti Jan. 2d» 1847 ; how long was it at interest ? OPERATION. . ri i 1847. 1st m. 2d day. Nofte.— In castinff interert, / 1846 12 ** 28 ** '''^ ^ finding the dinerence of Am, time between dates, each mondi is reclamed 90 days. 2. A note, bearing date Odt. 2Dth, 1823, Was paid April 25th, 1825 ; how long was the note at interest ? 3. What is the di&r^ce of time from Sept. 29th, 1844, to April 2d, 1847 ? Am, 2 y. 6 m. 3 d. 4. A man bought a farm Agril 14^h, 1842, and was to pay for it Sept. 1st, 1847, paying interest after Oct. 3()th, 1843 ; how much time had he in which to pay for the farm ? How- much time without ioterest ? For how long a time was he to pay interest ? ^ 5 yr. 4 m. 17 d. Am. M " 6 « 16 « ( 3 «* 10 " 1 « 1 1 i i Subtraction (rf Fraotionai Compomid Numbws. IT 198. From f of a week take H #^ ^ ^^7* Solution. — First seduce each fraetion to its ^per quantity, f of a week a= 2 da. 10 h. 12 m. ; ^ of a day = 10 h. 30 m. ; and 3 da. 19 h. 12 m.-«16 h. 30 m. =s 2 da. 2 h. 42 m. ^ An9. 2 da. 2 h. 42 m. Note. —We may, if we please, reduce the fractions to the same denomination, subtract them, and reduce their fractional difference to its proper quantity. From f of an ounce, take J of a pwt. Am. 11 pwt. 3 grs. From If of a hushel, take -ff of a peck. Am, 2 pk. 5 qts. 1 pt. From 2^ of a mile, take ^ of a furlong. Ans, 6 rds. 11 fL From \ of 19| gallons, take J of 3f quarts. Atu. 14 gal 3 qts. 1 pt. If gi. Questions, — % iST. How is the distance of time from one date to another tbun 1 ? 1 month is reckoned how many days ? t 1tt9. OOiyHICIND MUmiiML 1M» M«lt^ii0Kllan and Dtrteton of OodqMKUid ^umbers. 7139, To multiply and dhMe by 12 or lea. 1. A man has 3 pieces of doth, «ach measiurkiff fO yds. 3 qrs. ; how many yds. m the whole? 3. A man has 5 bottles, etch containing 2 gal. 1 qt. 1 pt ; what do they Mlcontain ? Heace, Compound IfoltipH* ^ctdon is, repeating a com- pouBd number as many times ai there arenuits in the mol* tiplier. & At I£. Ss. 8}d. per yard, what will 6 yards of clo&cost? 2. If 3 pieces of cloth con- tain 32 yds. 1 qr., how many yards in one piece ? 4. A man would put 11 gal. 3 qts. 1 jpt of Tinegar into 6 belles of the same size, what does each contain ? Hence, Compound Division iff, diriding a compound num- ber into as many parts as are indicated by the oiyisor. OPERATION. £. 9. d. qrt. 6)7 14 4 2 cost of 6 yards. 6. If 6 yards of cloth cost 7£. 14s. 4^d., what is the price per yard ? As the numbers are large, we write them down before mul- tiplying and dividing* OPEBATION. £. «. d. qrt. 15 8 S price of I yard. 6 number of yds. in*. 7 14 4 2 cost of 6 yards. Solution. — 6 times 3qrs. are 18qrs. = 4d. and Sqrs. over;, we write dowD the Sqrs. ; then, 6 times 8d. are 48d., and 4 to car- ry makes 52d.s4s. and 4d. over ; we "'write down the 4d. ; ^in, 6 times 58. are 30s. and 4 to carry makes 34s. = l£, and 148. over; 6 times 1J&. are 8£., and 1 to carry makes 7£.j which we write down ; and it is plain, that the united products arising from the several denomi- nations is the real product arising finm the whole cempour d number. 1 6 8 3 price cf 1 yard. Solution. — We divide *t£. by 6, to see how many £ each yard will cost, and find it to be IJ^. ; but 6 yards at \£, per 3rard would cost only 6 jE^. Henoe, the 6 yards cost \£, 14s. 4d. 2 qrs. more. Reducing \£. to shillings, and adding 14 shillings, we have 34 shillings, which, di- vided by 6, will give 5 shillings more as the price of each yard, and 4 shillings more, whieh, re* duced to pence and added to 4d. will make 52d., and dividing by 6, we have 8d. and 4 remainder ; this remainder reduced to qrs. is 16 and 2 are 18qrs., dividing by 6, the quotient is Sqis. Henceveaoh yard will cost \£. 58. 8d. 3<^nk i«e QOUPOmU) NUMBl^EUS. M29. The processes in the foregoing examples may now be pre- sented in form of a RIJUS. For mvltiplying a Com' pound Number when the mul^ tiplier does not exceed 12. Multiply each denomination separately, beginning at the least, as in multiplication of simple numbers, and carry as in addition of compound num- bers, setting down the whole product of the highest denomi- nation. For dividing a Compound Number when the divisor does not exceed 12. By short division, find how many times the divisor is con- tained in the highest denomi- nation, under which write the 'quotient, and if Qiere be a remainder reduce it to the jiext less denomination, add- ing thereto the number given» if any, of thai denomination, and divide as before. Proceed in this manner through all the denominations, and the several quotients will be the answer required. BXAMPUCS FOR PRACTICE. 7. What will be the cost of 5 pairs of shoes, at 10s. 6d. a pair ? 9. In 5 barrels of wheat, each containing 2 bu. 3 pks. 6 qts., how many bushels ? 11. How many yards of cloth will be required for 9 coats, allowing 4 yds. 1 qr. 3 na. to each ? 13. In 7 bottles of wine, each containing 2 qts. 1 pt 3 gills, how many gallons ? a At 2£. 12s. 6d. for 6 pairs of shoes, what is that a pair? 10. \i 14 bu. 2 pks. 6 ats. of wheat be equally divided into five barrels, how many bushels will each contain ? 12. If 9 coats contain 39 yds. 3 qrs. 3 na., what does 1 coat contain ? 14. If 5 gal. 1 gill of wine be divided equally into 7 bot- tles, how much will each con- tain? Qnestions, — If 129* What is compoond multiplication ? Where do you begin to multiply ? How do you proceed when the multiplier does not exceed 12? How do you carry? Repeat the rule, when the multiplier is 12 or less. What is compound division? How do you write the numbers when the divisor does not exceed 12? Where do you begin the division? Where write the quotient ? If there be a re- maind«r» how do you proceed ? Repeat the rule. 1130. COMPOOND NCMBBBa 167 I 15. What will be the weight of 8 silver cups, each weighing 6 oz. 12 pwt. 17 grs. ? 17. How much sugar in 12 hogsheads, each containing 9 cwt. 3 qrs. 21 lb. I 19. How much beer in 9 casks, each containing 1 bar. 7 gal. 3 qts. 1 pt. ? 21. A house has 7 rooms, averaging 1 sq. rod 57 sq. ft. 55 sq. inches ; what do they all contain ? 23: A boat on the Erie canal averages 21 m. 65 rods 13 ft. a day ; what is that for 5 days ? 25. What quantity of land in 6 fields, each containing 17 A. 7 sq. C. 12 sq. rd. 133 sq. 1.? 27. How much wood in 12 piles, each containing 7 C. 5 c. ft. 12 cu. ft. ? IT ISO. To multiply and 1. In 15 loads of oats, each measuring 42 bu. 3 pk. 2 qts., how many bushels ? Solution. — Multiplying the quantity in 1 load by 3, we have the quantity in 3 loads, and mul- tipljdng the quantity in 3 loads by 5, we have the quantity in 5 times 3, or 15 loads. 16. If 8 silver cups weigh 3 lb. 9 oz. 1 pwt. 16 m., what 18 the weight of each ? 18. If 119 cwt. 1 qr. of su- firar be divided into 12 hogs- heads, how much will each hogshead contain ? 20. If 9 equal casks of beer contain 10 bar. 34 gal. 3 qts. 1 pt., what quantity in each? 22. If 7 rooms contain 8 sq. rods 129 sq. ft. 61 sq. in., what is the average ? 24. A boat moves 106 m. 8 rods 15 ft. 6 in. in 5 days, what is the average of each day? 26. If 6 equal fields contain 106 A. 6 sq. C. 9 sq. rds. 173 sq. 1. ; how much land in each? 28. In 12 piles of wood are 92 C. 5 c. ft., how much in each? divide by a composite number. 2. If 15 loads of oats meas- ure 642 bu. pk. 6 qts., how many bushels in each load ? Solution. — Dividing the quantity in 15 loads by 5, we have the quantity in -^ of 15, or 3 loads, and dividing the quantity in 3 loads by 3, we have the qi&ntity in 1 load. QnestiOBS. — 1 130, VH^*^ the multiplier, or divisor, is a com- posite number, how may the operation be contracted in multiplicalion? —in division? I« OOMPOtJNl^ iTtTMfiERS. Tldl OPEBATION. bu. pk.qta. 42 3 2 m 1 load. 3 one factor. One^ OPERATION. bu, pk. ft». 128 I 6 in 3 loads. 6 the oiher factor. j^^{$)e^ 6inl5loadM ^2;:-^3)128 1 QmSloais. Ans. ^ S 2inl load. 642 6 m 15 loads, Ans. Hence, W?ien the multiplier or divisor exceeds 12 and is a wnvposite immher, RUUB. Multiply by each of the component parts of the multi- plier. The last product will be the answer. 3. What will 24 barrels of flour cost, at 2£. 12s. 4d. a barrel? 5. How many bushels of apples in 112 barrels, each barrel containing 2 bu. 1 pk. ? Note. — 8, 7, and 2, are factors of 113. 7. How much molasses in 84 hogsheads, each hogshead containing 112 gal. 2 qts. 1 pt. 3 gi. ? 9. How many bushels of wheat in 135 bags, each con- taining 2 bu. 3 pecks ? 3X9X5 = 135. 11. Sold 25 pieces of cloth, each containing 32 yds. 2 qr. 1 na. ; how much in the whole ? Divide by each of the com- ponent parts of the divisor. The last quotient will be the answer. 4. Bought 24 barrels of flour, for 62£. 16s. ; how much was that per barrel ? 6. In 112 barrels are 252 bushels of apples ; how many- bushels in 1 banrel ? 8. Bought 84 hogsheads of molasses, containing 9468 gal. 1 qt. 1 pt. ; how, much in a hogshead ? 10. 371 bu. Ipk. of wheat are equally divided into 135 bags ; how much in each ? 12. Sold 814 yds. 1 na. of cloth in 25 equal pieces ; how much in each piece ? IT 131 • To muLti'ply arid divide hy any numher greater than 12, which is not a composite number. 1. How many yards of 2. Bought 139 pieces of sheeting in 139 pieces, eaor sheeting, containing 4439 yds. Siece containing 31 yds. 3 qrs. 1 qr. 1 na. ; how many yards na. ? in 1 piece ? TI31. COMBDUND NUMBERS. 169 Solution. — 139 is not a com- posite number. We may, how- ever, decompose this number thnSy I39=10ft-f 30 + 9. We may now multiply the num- ber of yards in 1 piece by 10, which will give the numl>er of yards in 10 pieces, and this prod- net again by 10, which will give the number of yards in 100 pieces. We may next multiply the number of yards in 10 pieces by 3, which will give the number of yards in 30 pieces, and the number of yards in 1 piece by 9, which will give the number of yards in 9 pieces, and these three products, added together, will evidently give the number of yards in 139 pieces; thus: yds. 31 qrs. 3 na. 3 in 1 piece. 10 319 1 2 in 10 pieces. 10 3193 958 287 3 1 in 100 pieces. 2 in 30 pieces. S in 9 pieces. 4439 1 1 in 139 pieces. Note. — In multiplying the ttumber of yards in 10 pieces, (319 yds. 1 qr. 2 na.,) by 3, to get the number of yards in 30 pieces, and in multiplying the number of yards in 1 piece, (31 yds. 3 qrs. 3 na.,) hj 9, to get the number of yards m 9 pieces, the multipliers, 3 and 9, need not be written down. Solution. — When the divisor cannot be produced by the multi- plication of small numbers, the Detter way is to divide after the manner of long division, setting down the work of dividing and re- ducing as follows : yds. qr. na. yds. qr. nt 139)4439 1 1(31 3 3 417 269 139 130 4 621(3 qrs. 417 104 4 417(3 na. 417 We divide 4439 yds. by 139, to ascertain the number of yards in each piece, which we find to be 31, and a remainder of 130 yds., which will not be a whole yard to eacli piece ; but reducing them to quarters, adding in the 1 qr., we have 521 qrs., which we divide by 139 to ascertain the number of quarters in each piece, and find it to be. 3, and a remainder of 104 qrs. ; we reduce this remainder to nails, adding in the 1 nail, and have 417 na., which we divide by 139 for the number of nails in each piece, and find it to be 3, without a remainder. Hence, there are 31 yds. 3 qrs. 3 na. in each piece, Ans. Qvestions, — IT 131. When the multiplier or divisor exceeds 12, how is the multiplication performed ? — the division ? 15 170 QOWPOVND NUipEBa TIM Hence, When the multiplier or divisor exceeds \% amd is Tiot a composite nurnbert RUUB. Multiply first by 10, and Ais product by 10, which will give the product for 100 ; and if the hundreds in the multi- plier be more than one, multi- ply the product of 100 by the mimher of hundreds ; for the tens, multiply the product of 10 by the number of tens ; for the units, multiply the multi- plicand; and the sum of these several products will be the product required. Divide afler the manner of long division, setting dowa the work, of dividing and reducing. EXAMPLES FOR PRACTICE. 3. How many acres in 241 wild lots, each containing 75 acres 2 roods 25 rods ? 4. There are surveyed in an unsettled district, 18233 acres 25 rods of land, which is divided into 241 equal lots ; how many acres in each lot ? 6. If 1806 lbs. 15 oz. of tea be divided equally into 23 chests, how much will be in ea'ch chest ? 5. How many pounds of tea in 23 chests, each contain- ing 78 lbs. 9 oz. ? Note. — The pupil will easily perceive the method of operation, when the multiplier is less than 100. 7. Bought 375 bales of 8. Bought 375 bales of English goods at 9£. lis. 6d. English goods for 3590£. 12s. per bale ; what did the whole 6d. ; what did each bale cost ? cost? 9. How many bushels of 10. A wealthy farmer har- wheat are raised on 125 acres, vested 125 acres of wheat, averaging 22 bush. 3 pecks 5 which yielded 2863 bush. 1 quarts to the acre ? peck 1 qt. ; what was the average per acre ? TISe. COKPOUNP HUHBEBS 171 placa. Every circle, whether great or small, is supposed to be (ii> filled into 360 equal pans, called degrees. Let the accompanying dia- gram repreient the great cir- cle of the earth, called th« eqttaloT, divided, as you he e see, into 24 equal parts of 15 degrees each, 136&' -^ 24 = 15».) As the sun apparently passes round the earth in 24 hours, it will pass through one of tjiese divisions, or 15", in 1 hour ^60 minutes of time, and of course it will pass 1° of motioii in (l^ of 60^4 minutes of time, and 1' of motion in ^ of 4 minutes (= 240 seconds -i- 60) ^ 4 seconds of time. Hence it follows, that the apparent motion of the sua round the earth, from east to west, is 15° of motion in 1 hour of time, 1" of motion in 4 minutes of time, and 1' of motion in 4 seconds of time. From these premises, it follows that when there is a difference in long!- there is a difference in time tude tetween two places there between two places, there is a will be a corresponding differ- corresponding difference in ence in the hour, or time of their longitude. If the differ- the day. The difference in ence in time be 1 hour, the longitude being' 15°, the dif- difference in longitude will be ference in time will be 1 hour; 15° ; if 4 minutes, the differ- the place etu^erZy having noon, ence in longitude will be 1", or any other specified time, 1 &c. hour sooner than the place Hence, if the difference in westerly. time (in minutes and seconds) Hence, if the difference in between two places be divided longitude, in degrees, and by 4, the quotient will be the minutes, be multiplied by 4, difference in hngitude, in de- the product will be the differ- grees and minutes. ence in time in minutes and seconds, which may be re- duced to hours L^iL. 172 COMPOUND NUMBERS. T133 2. What is the difference in longitude between London and Washington, the differ- ence in time being 5 h. 8 m. ? Solution. — The sun's appar- ent motion is 1° in 4 minutes of time, and we wish to know how far it will move in 308 minutes. It will move as many degrees as 4 min. is contained times in (can be subtracted from) 308 min And 308 -^ 4 = 77^, Ans. 4. When it is 12 o'clock at the most easterly extremity of the island of Cuba, it is 16 minutes past 11 o'clock at the western extremity; what is the difference in longitude be- tween the two points? 5. Supposing a meteor should appear so high that it could be seen at once by the inhabitants of Boston, 71® 3', of Wash- ington, 77° 43', and of the Sandwich Islands, 155® west lon- gitude ; if the time be 47 minutes past 11 o'clock of Dec. 31, 1847, at Washington, what^will be the time at Boston, and at the Sandwich Islands ? ' Ans. At Boston, 13 min. 40 sec. A. M. (morning) of Jan. 1st, 1848 ; at the Sandwich Islands, 37 min. 52 sec. past 6 o'clock, P. M., of Dec. 31, 1847. 1. What is the difference in time between London and Washington, the difference in longitude being '77** ? Solution. — The sun appar- ently moves 1° in 4 minutes of time, and it will move 77° in 77 times 4 minutes = 308 minutes. Or, we may multiply 77 by 4, since either factor may be the multiplicand. And 308 min. =s 6 h. 8 min, Ans. 3. When it is 12 o'clock at the most easterly extremity of the island of Cuba, what will be the hour at the most wes- terly extremity, the difference in longitude being 11® ? IT 133. Review of Comi>ound Numbers. Qacstions* — What distinction do you make between simple and compound numbers? rj[ 102.) What is the rule for adding con^und numbers ? In ^ 124, JEx. 33, what difficulty is met with in carrying from miles to degrees ? How is it obviated ? Rule for subtracting com- ' V ' — Qaestions. — ^ 132* What circle is the diagram intended to rep- resent ? Into how many divisions is it divided ? how many degrees in each division ; and what does it represent ? In what time does the sun move 1°, and why ? — 1', and why ? What does M.*ignify ? AM.? P. M.? What motion causes the apparent motion of the sun ? When the difference of longitude between places is known, how may the differ- ence in time be calculated ? When it is noon at any place, is it before noon or after noon at places easterly ? — at places westerly ? Why ? When there is a difference in. time between places, what follows? When the difference in time is given, how may tne diierence in longi- tude be found? T 133. COMPOUND NUMBERS. 173 poand numbers? — for moltiplying when the mnltipIieT does not exceed 12 ? — when it does exceed 12, ana is a composite number ? — w'.ien not a composite number ? — for dividing compound numbers, when the divisor does not exceed 12? — when it exceeds 12 and is a composite number? — when not a composite number ? How is the distance of time from one date to another found ? How many degrees does the earth revolve from west to east in 1 hour ? In what time does it revolve 1° ? Where is the time or hour of the day sooner — at the place most easterly or most westerly ? The difference in longitude between two places being known, how is the difference in time calculated ? The difference in time being known, how is the difference in longitude calculated? BXSRCISBS. 1. A gentleman is possessed of 1^ dozen of silver spoons, each weighing 3 oz. 5 pwt. ; 2 doz. of tea-spoons, each weigh- ing 15 pwt. 14 gr. ; 3 sliver cuns, each 9 oz. 7 pwt. ; 2 silver tankards, each 21 oz. 15 pwt. ; and 6 silver porringers, each 11 oz. 18 pwt. ; what is the weight of the whole ? Atis, 18 lb. 4 oz. 3 pwt. 2. In 35 pieces of cloth, each measuring 27 yds. 3 qrs., how many yards ? Atis, 971 yds. 1 qr. 3. How ;iiuch wine in 9 casks, each containing 45 gal. 3 qts. 1 pt. ? 4. If a horse travel a mile in 12 min. 16 sec, in what time would he travel 176 miles ? il?M. Id. 11 h. 58 m. 56 sec. 5. If 8 horses consume 889 bu. 2 pks. 6 qts. of oats in 365 days, how much will one horse consume in 1 day ? Atis, 1 pk. 1 qt. 1 pt. 2 gills. 6. I hold an obligation against George Brown, of London, of 735£. lis. 6d., on which are two endorsements, viz., 61£. 5s. and 195£. 13s. lid. ; what remains unpaid? Am, 478£. 12s. 7d. 7. Liverpool, Jan. 1, 1848. Isaac Derwent, of Boston, U. S. Bought of Shipley & Co. 10 boy's hats. No. 1, 4s. 6d. 12 do. 2, 4 do. 3, 4 do. 9, 4 d:>. "* 10, 6 do. 11, 6 men's hats, 1 trunk for packing, Received payment, 15* 5s. 5s. 6d. 10s. lis. 12s. 14s. 1£. 4s. 19£. lis. Shipley & Co. by G. Willirims. •l.B'W*^. ti., ^ ". J«.. t''\ Ml ' ' " '^ •. Jtf J'-, 174 ANALYSIS. H 1J34. Kt)TE. — If the chr5e foHo'mng examples be wiought correctly, flie Answers will be ^ here given. 8. A mail dividss 16 bar. 23 gal. 3 qts. of oil into 5 large vessels ; how much does he put in each ? Am. 3 bar. 11 gal. 0| qt. 9. On an acre of ground were erected 21 buildings, occu- pying on an average 3 sq. rds. 112 ft. 81 in. ; how much re- mained unoccupied ? Ans. 2 roods 8 rods 86 ft. 63 in. 10.' If a man build 3 rds. 9 ft. 7 in. in length of wall in 1 day, how many rods can he build in 15 days ? Ans. 53 rds. 11 ft. 9 in. 11. If a ship sail 3^ 18' 45" in one day, how far will it uail in the month of June ? Ans. 99° 22' 30". 12. If a druggist sell 1 gross 7 doz. bottles of sarsaparilla in 1 week, how many gross will he sell in the months of April, May and June, at the same rate ? Ans. 20 gr. 7 doz. 13. If a man, employed in counting money from k heap, count 100 silver dollars in a minute, and continue at the work 10 hours each day, how many days will it take him to count a million i Am. i6| days. 14. At the same rate, how many years, reckoning 366 days to a year, would it take him to count a billion ? Ans. 45 years 241| days. 15. "Were 1000 men employed at this same business, each one counting at the same rate, 10 hours each day, how many years would it take them to couiit a quadrillion ? Ans. ^662 years 36f days. ANALYSIS. V li4. In most examples in arithmetic, two things are given to find a third. Thus, in the relations of the price and quantity, the quantity and the price of a unit may be given to find the price of the quantity, or the quantity and its price to find the price of a unit, or the price of a unit and of a quan- tity to find the quantity. The same principles may readily be applied to other calculations. This method of operating is called Analysis. Analysis, therefore, may be defined, the solving of questions on general principles. We have presented (IF 46) these rules in connection, as ap- plied to Whole numbers, and separately as applied to any T134 ANALTSia 175 quantities in TT 77, 86, and 83. We will now present them together as applied to any quantities, with examples which mutually prove each other, except where there are some frac- tional losses. I. The price of vnity, artd the quanr tity betTtg given, to fnd the price of the qmniity^ n. The quanti' III. The price of ty, and the price of unity ^ and the price Vie quantity being of a quantity being given, td find the given, to find the price of unity y quantity. Multiply the price ky the quantity. RUUS. Divide the cost by the quantity. BXAMPIiSS FOR PRACTICSU RUIiB. Divide the price of the quantity by the price of unity. 1. At $302*40 per tun, what will Ihhd. 15gal. 3qts. ofwme cost? , 4. At $2*215 per gal., what coSt 34 qts. ? ^ o ; - 7. At ^6*72 per ton for pot-ashes, whatwillf of aton cost? 10. If a yard of cloth cost $2*5, what will *8.of a yard cost ? 13. If a ton of pot-ashes cost £27 10s., what will 14 cwt. cost ? 16. If a bushel of wheat cost Sl'92, what will 1 pk. 4 qtb cost ? 2. At $94*50 for 1 hhd. 15 gal. 3 qts. of wme, what is that per tun? / 5. At $1*80 for 3i qts. of wine, what is that per gal.? 8. If f of a ton of pot-ashes cost $60*45, what is that per ton ? 11. If *8 of a yard of cloth cost $2, what is that per yard ? 14. If 14 cwt. of pot-ashes cost £19 5s., what is that per ton ? 17. If 1 pk. 4 qts. of wheat cost $*72,.what is that per bushe ? 3. At $302*40 per tun, how much wine may be bought for $94*50. 6. At $2*215 per gal., how much wine may be bought for $1*80 ? 9. At $96*72 per ton, how much pot- ashes may be bought for $60*45 ? 12. At $2*5 per yard, how much cloth may be pur- chased for $2 ? 15. At £27 10s. a ton for pot-ashes, what quantity may be bought for £19 5s.? la At $1*92 per bushel, how much wheat may be bought for $*72? 176 ANALYSIS. T134w 19. If a yard of broadcloth cost $6, what will 16 yds. 2 qrs. 3 na. cost ? 22. If a ton of hay cost $13, what will 1850 lbs. cost? 25. If an eagle weigh 11 pwt. 6 grs., what will be the weight of 665 eagles ? 20. If 16 yds. 2 qrs. 3 na. of broad- cloth cost $100* 125, wliat is that per yard? 23. If 1850 lbs. of hay cost $12- <025, what is that per ton ? 26. If665eagles weigh 31 lbs. 2 oz. 1 pwt. 6 grs., what is the weight of each? 21. At $6 per yard, how much broadcloth may be bought for $100- •125? 24. At $13 per ton, how many lbs* of hay may b^ bought for $12* *025? 27. How many eagles, each weigh- ing 11 pwt. 6 grs., may be coined from 31 lbs. 2 oz. 1 pwt. 6 grs. of standard gold? Note. — After the same manner let the pupil reverse the follow ing examples : 28. At $1*75 a bushel for wheat, how many quarters caj» be bought for $200 ? Am. 11 qrs. 2 bu. 1 pk. ia428f qts. 29. What will 3 qrs. 2 na. of broadcloth cost, at $6 pe yard ? * 30. At $22*10 for the transportation of 6500 lbs. 46 miles* what is that per ton ? $6*80. 31. Bought a silv' r cup, weighing 9'OZ. 4 pwt. 16 grs., fw $1 1*08 ; what was t) at per ounce ? 32# A lady purchased a gold ring, giving at the rate o^ $20 per ounce ; she paid for the ring $1*25 ; how much d>A it weigh ? Am. 1 pwt. 6 grs. Examples requiring several operations. 33. If 6 bushels of wheat cost $7*50, what will 28 bushels cost ? Note !• It was customary formerly to perform all examples suni- Qaestions. — ^ 134» What do you say of every example in arith- metic ? What severally are given and required in the relations of the price and quantity ? Answer the questions in ^ 46. {The teacher rviV, ask them.) What is given and what required under I. ? — the rule ? — under II. ? — the rule ? — under III. ? — the rule ? How were such examples as the 33d formerly wrought ? What method is preferable ? What do we do by this method ? Explain the solution of this example : — the first solu- tion of example 34 1 — - the second method. What is this called ? What is an analytic solution ? 1 134. ANALYSIS. 177 hi to the above by a rale caUed the Rule of Three j or, on aociiunt of its supposed importance, the golden rule. But they are more intelligi- bly solved by analysis ; that is, from the things given in the question, find the price of unity, and having the price of unity, find the price of the quantity, the pric6 of which is required. Thus, OPERATION. 6)$7'50,pric€of6hu. $1*25, " 1 *t*. -J Tl M- *i. • ^ « OQ Solution. — Dividing the price of 6 bu. by 6 wiU give the price of 1 ba., -^^^ and multiplying the price of 1 bu. by l^)!y 88 will give the price of 28 bu. 250 at $35*00, price of 28 lu. 34^ li\^oi9^ bushel of com cost -^ of a dollar, what wiD f |- of a bushel cost ? Solution. — We divide the price of the quantity, -f^, by the quantity, -fj, to get the price of unity : -/^ -5- -}-J=-^^ of a dol- lar, price of 1 bu. Ajid multiplying the price of 1 bu. by Jf of a bushel, will give the price of the quantity required : -^^ X Jf ^5= ^^ of a dollar. Or, to give a more full analysis, if -j^ of a bushel cost -j^g^ of a dollar, -j^ij- would cost -j^ as much : .jij- of -jj^ = y|-j ; and \^ = 1 bu., would cost 13 times as much as -j^. y^-g- X 13 = x^V ^^ a doUai^, price of 1 bushel. Then -^ of -^^=zj^-^y is the price of ^ of a bushel, and 32 times the price of 5^, that is, y%^ X 32 ^z^^ of a dollar, is the price of f f of a bushel. Note 2. This process is called an analytical solution, or solving the question by analysis. 35. If 16 men will finish a piece of work in 28J days, how •ong will it take 12 men to do the same work ? Note 3. Find in what time 1 man would do it, and 12 men would do it in -j^ of the time. Ans, 37|^ days. 36. How many yards of alpacca, which is 1^ yards wide, will be required to Kne 20 yards of cassimere ^ of a yard wide ? Note 4. Find the square contents of the cassimere, which is the same as the square contents of the alpacca. We have, then, the square contents and the width of the alpacca given to find the length? % 49. Ans, 12 yds. of alpacca. 1, \ 17i PRACtlCE. i iS5 d7. If 7 horses consume 2| tons of hay in 6 weeks, hdW much will 12 horses consume m 8 weeks ? Ans, 6f tons. 38. If 5 persons drink 7^ gallons of beer in 3 week, how much will 8 persons drink in 22^ weeks ? Ans. 280f gallons. 39. A merchant bought several bales of velvety each con- taining 12944- yards, at the rate of $7 for 5 yards, and sold them at $11 for 7 yards, gaining $200 on them,^ how many bales Were there ? "^^^ * Ans. 9 bales. 40. If ^ lb. less by | costs 1?^., what cost 14 lb. less by I of 2 lb. ? . Ans. £4 9s. 9^. 41. If 5 acres 1 rood produce 26 quarters 2 bushels of wheat, how many acres will be required to produce 47 quar- ters 4 bushels ? ^ Am, 9 A. 2 K. 42. If 9 students spend £10|- in 18 days, how much will SO students spend in 30 days ? Ans. £39 18b. 4|fd. 43. If f yd. cost $J, what will 40^ yds. cost ? Ans. $59*062 4-. 44. If -^ of a ship costs $251, what is ^ of it worth ) Ans. $53*785 -|-. 45. At £3| per cwt., what will 9 J lbs. cost ? Ans. 6s. 3-^d. •it i 46. A merchant, owning f of a vessel, sold f of his share ! for $957 ; what was the vessel worth ? Ans. $1794*375. 47. If f yd. cost £f , what will -^ of an ell Eng. cost ? J . J _ c. -^^' I'^s. Id. 2fq. ^ ■ — 7 ;r— r '. ■ ■'., -■ f PRACTICE.' '.' • -. /■'l.'t.-i- />-/ -^ • / ^; IT 13ff« I. TT^ the pria^ is an dliqtiot part of a de6ar. Note. — For the definition of aliquot part, see ^ 55. ALIQUOT PARTS OF 1 DOLLAB. Cents. Cents. 50 = J of 1 dollar. 12 J = ^ of 1 dollar. ?3 J = ^ of 1 dollar. 10 = ^^ of 1 dollar. 25 = I of 1 dollar. 6^ = ^^ of 1 dollar. 20 =i of 1 dollar. 5 =2VoflddUar. 1. What will be the cost of 4857 yaids of catice at SIS tents (a« J of 1 dollar) per yard ? \ f 136. PRACTICE. • 179 OPERATION. Solution. — \i $1 a 3rard, the <^ would 4)4857 dollars, ^ ^ many doUan as thdre are yards, that is, ! $4857 ; and at 4 cf a dollar a yard, it is plain, \ $1214*25, Ans, that the cost will be ^ as many dollars as there t are yards, that is, •a]^ = f 1214'25. { Ailer dividing the miit figure 7, there is a remainder of 1, (dollar,^ \ which we reduce to cents by annexing ciphers, and continue the divi \ sion. This manner of computing the cost of articles, by taking aliquot parts^ is called rRACTiCE, from its daily use among merchants and tradesmen. Hence, when the price is an aliquot part of a dollar, this general RUIiE OF PRACTICE. Divide the price at $1 per pound, yurd, &c., by the num- ber expressing -the aliquot part, the quotient will be the an- swer in dollars. Note. — If there be a remainder, it may be reduced to cents and mills, by annexing ciphers, and the division continued. SXAMPIiES FOR PRACTICE* 2. What is the vallie of 14756 yards of cotton cloth, at 12J cents, or I of a dollar per yard ? By practice. By mvltiplicaticm. Note. — By 8) 14756 14756 comparing the i\2^ two operations. An.. $1844.50 _— If L^^ YdYoU jjqh \^y practice 29512 is much shorter 14756 than the one by — multiplication. $1844*500 Ans. as before. 3. What is the cost of 18745 pounds of tea, at $*50, = \ dollar, jer pound ? Ans. $9372*50. 4. What is the value of 9366 bushels of potatoes, at 33J cents, or j of a dollar, per bushel ? A4p=$3122, Ans. *■ , Qaestions* — If 135. What do you understand by aliquot parts? What are the aliquot parts of a dollar 1 When the price of 1 yard, 1 pound, &c., is an aliquot part of a dollar, how may the cost of any ^antity of that article be found? What is this manner of computing called ? Why ? Repeat the rule. What two things are g'ven, and what one is required, in example 3d? — in example 4th? 5th? 0th? 7th I 4cc. i' 180 • PRACTICE. IT 196. 5. What is the value of 48240 pounds of cheese, at $*06^, = -j^ ofa dollar, per pound ? Aits, $3015. 6. What cost 4870 oranges, at 5 cents, = ^ of a dollar, apiece ? Am, $243*50. 7. What is the value of 151020 bushels of apples, at 20 cents, = i of a dollar, per bushel ? Am, $30204. 8. What will 264 pounds of butter cost, at 12i cents per pound ? Am, $33. 9. What cost 3740 yards of cloth, at $1*25 per yard ? 4 ) $3740 = cost at $1* per yard. 935 = cost at $ *25 per yard. Am, $4675 = cost at $1*25 yer pard. 10. What is the cost of 8460 hats, at $1*12 J apiece ? at $1*50 apiece ? at $3'20 apiece ? at $4*06^ apiece ? Am, $9517*50. $12690. $27072. $34368*75. IT 136. U' To find the vcHue of articles sold by the 100, or 1000. 1. What is the value of 865 feet of timber, at $5 per hun- dred^ Were the price $5 per foot, it OPERATION. is plain the value would be 865 ggg ' X S5 = $4325 ; but the price is e $5 for 100 feet; consequently, ^ $4325 is 100 times the true value "T Oh IT /• of the timber; and therefore, if $4325 = value at $5 per foot, we divide this number ($4325) by 100, we shall obtain the true value ; which we do by cutting off two right hand figures. Ans. $43*25. Were the price so much per thousand, the same remarks would apply, with the exception of cutting off three figures, instead of two. Hence, To find the value of articles sold by the 100 or 1000, I. Multiply the number and price together. II. If the price be by the 100, cut off two figures at the right ; if by the 1000, cut off three figures at the right ; the prodiict will be the answer, in the same denomination as the price, which, if cents or mills, may be reduced to dollars. ^ T 137. PRACTICB. 181 2. What is the cost of 4250 bricks, at $5<75 per 1000 ? Ans. $24<43|. OPERATION. $5*75 4250 In his example, we cut off three figures from the rieht hand of the product, because the bricks were 2&75 ^^ by the 1000. The remaining figures at the left ^ . -^ express the cost of the bricks in the lowest den4imi- OQHA nation of the price, viz., cents, which we reduce to 2300 dollars by pointing off two places foi cents $24*431750 , 3. What will 3460 feet of timber cost, at $4 per h dred? 3 ^ ^/^ 4. What will 24650 bricks cost, at 5 dollars per 1000?, ^ 6. What will 4750 feet of boards cost, at $12*25 pel 1000 ? Ans. $58*187+. 6. What will 38600 bricks cost, at $4*75 per 1000 ? / f j 7. What will 46590 feet of boards cost, at $10*625 per 1000? ^9j-'(jJ/-f V 8. What will 75 feet of timber cost, at $4 per 100 ? ^ ' ^ 9. What is the value of 4000 bricks, at 3 dollars per 1000 ? Wilderness, February 8, 1847. ter, Bought of Asa Falltree, at $6 per M. " 12*64 ." (( 4 (( « 10 " " 2*50 per C. " 2*75 ** / * > Ml •. Peter Carp( 1 5682 feet Boards, 2000 i( 800 1500 879 Thick Stuff, Lathing, Plank, Timber, 236 (( Received payment, $101*849. Asa Falltree. Note. — M. stands for the Latin milk, which signifies 1000, and C. for the Latin word centuniy which signifies 100. IT 1 37, IIL To find the cost of articles by the ton of 2000 lbs. 1. What cost 3684 lbs. of hay, at $12*40 a ton ? _,!--■ ■ I - !■ ■ - - ^- ■ I I II - ■ I ■ I I I I I --|-| r ^T Qnestions. — % 136« To find the cost of articles sold by 100, or i.000, what is the first step proposed by the rule? — the second? In what denomination will the produt : be? How will you find the cost of 725 tnicks, at $4'25 a thousand ? How many figures in all do we point off f 16 ■] f n cuui: 1^ PRACTICE. tl3t. OPERATION. SoLUm^N. — A ton J12»40 -?- 2=± $6*20= ©rtcc of 1000 Z3* equals 2000 lbs. Di- ^gtgo Tiding the price of one o/»rv4 ton by 2, we have the *^^^^ price of 1000 lbs., ^ZTZTTZ^ ' which is $6*20. Mul- $22*84 1 080 . tiplying this cost by the number of pounds, 8684, it will give the cost at $6^20 per pound, a result which is 1000 times too large. ^ 136* We therefore divide this product by 1000, cutting off three right hand figures, and have the cost at $6*20 per lbs., or $12*40 per ton. An«. $22*84+. ence, Multiply 1 half the cost of 1 ton by Jthe number of pounds, and point off three figures from the right hand. The remain- ing figures will be the price, in the denomination of the price of 1 ton, which, if cents or mills, may be reduced to dollars. Note. — At $12*40 per ton, or $6*20 per 1000 lbs. 100 lbs. will cost $ *62 removing the separatrix 1 figure to the left. 10 ** " $*062 " " 2 figures ** 1 *< " $*0062 ** ** 3 ** " 2. What is the cost of 15742 lbs. of Anthracite coal, at $7*50 per ton ? Am, $59*032-j-. 3. What will be the transportation on 49826 lbs. of iron, fcem New York to Chicago, at $11 per ton ? Am, $274*043. 4. What wiU be the storage on 13991 lbs. of goods, at $2*50 per ton ? Am, $17'488-|-. 5. What will be the cost of 658 lbs. of hay, at $7*38 per ton ? at $5*25? at $8*50? at $9*00? at $9*50? at $11 ? at $J2? Am, $2*428. $1*727. $2*7964. $2*961. $3* 125 J. $3*619. $3*948. 6. At $7*00 per ton, what will be the cost of 424 lbs. of hay? 530 lbs.? 658 lbs.? 750 lbs.? 896 lbs.? —^918 lbs.? 1024 lbs.? 1216 lbs.? 1350 lbs.? 1600 lbs.? 1890 lbs.? Am, $1*484. $1*85J. $2*303. $262J. $3*136. $3*2ia $3*584. $4*256. $4*72^. $5^60. $661^. Questions. — ^ 137. When the cost of 1 ton is given, how do you find the cost of 1000 pounds ? How do you find the cost of any number of pounds? i iSa PRACTICE. 183 V 13S. IV. When the price is the aliquot part of a £. JlLiqUOt PABTS OF A £. 10s. = I of 1£. 6s. 8d. = ( of 1£. 5s. = 1 «• 3s. 4d. = I " 4s. = I « 2s. 6d. => I " 2s.=tV " Is. 8d. = ^ " 1. Bought of John Smith, Liveipool, 7685 yards of black l)f oaddoth at lOs. per yard ; what (fid it cost ? bPERATioit. Solution. — At 1 i^ . pertr 2) 7685£. =:price at 1£. per yard. T^ the price is 7C86je., and one half of this is the OQ^of IOk =« 10* " price at IDs. per yard. The ,»4^. JW.= ws, remaiDder of li. may be xeduoed to diilHngB and then divided. HencO, when the price is the aliquot part of 1£., Divide the price of the quantity at 1£. per yard, bushel, &c., by the number expressing the aliquot part. The answer will be in pounds. mXAMTPliBS FOR PRACTICE. 2. What cost 1873 reams of paper, at 6s. 8d. per ream ? Am. 624£. 6s. 8d. 3. What cost 10416 bushels of salt, at 3s. 4d. per bushe. ? A71S, 1736£. 4. Bought 640 lbs. colored thread, at 7s. 6d. per pound ; what was the whole cost ? OPERATION. Solution. — 78. 6d. is net an ali- 814) 640£. =: 1£. ver Ih. ^^^^ P^^ of a pound, but it is equal to ' ' ^ 6e. + 3s. 6d., and taking k of 640jC. Qi \af\r y?c " we have the price at 5s. per lb., and ' on^' ~ o aj u * of 640 jC., or i of 160JC.,is the price OUi,. = ^s. btf. at 2s. 6d. Then, adding together the prices at 5s. and 2s. 6d. per lb., we 240£. = 75. M, " have the price at 7s. 6d. per lb. 5. What cost 866 yards of black silk, at 14s. per yd. ? Am. 606£. 4s. 6. What cost 7 T. 8 cwt. of iron, at 16s. 8d. per cwt. ? Am. 123£. 6s. 8d. C^nestions. — T138. Give the aliquot parts of 1£. Explain the principle on which the first example is performed? — the fourth exam* ^? Rule. *j.*..^ 184 PRACi ICE. T 139, 140. iri39* V. When the price is an aliqttot part of a shilling. ALIQUOT PARTS OP A SHILLING. 6d. = i of Is. 2d. = i of Is. 4d. = i " IJd. = i - 3d. = i " Id. =rA " Reasoning as above, we have this Divide the price of the quantity at Is. per lb., yd., &c.,'by the number expressing the aliquot part ; the answer will be in shillings. EXAMPUBS FOR PRACTICE. 1. Sold 348216 lbs. of cotton, for 4d. per lb. ; what did 1 receive? Ans. 116072s. =6803£. 12s. 2. Bought 2490 yds. of calico, at 9d. per yard ; what did it all cost ? 9d. =6d. + 3d. Ans. 93£. 7s. 6d. 3. Bought 4000 papers of pins, at 4^. per paper ; what was the cost ? 4Jd. =3d. -}- IJd. 4. What cost 7430 lbs. of sugar, at 6d. per lb. ? at 4d. ? at 3d. ? — :— at 2d. ? at IJd. ? at Id. ? IT 140. To Jind the price of a quantity less than unity, when it is an aliqicot part^ or par is y of\, . 1. At $1^50 per bushel for wheat, what will 2 pecks and 4 quarts cost ? ft^l nT^O^' Solution. — We divide the price of 1 bushel by 2, o\4) I o\j which gives the price of 2 pecks, or i a bushel. Then, as 4 quarts is ^ of 1 bushel, we divide $1*50 by 8, or 4 ) *75 as it is k of half a bushel, we divide $*75 by 4, for the *18| price of 4 quarts, and the quotient, added to the price of 2 pecks, gives the price of 2 pecks and 4 quarts. (93| Ans, $*93i. Hence, Take such part, or parts, of the price of unity as the quan- tity is of 1 ; the part, or sum of the parts taken, will be the price of the quantity. Qnestions, — ^ 139. Give the aliquot parts of 1 shilling. Rule Show how the 2d example can be performed by Practice j the 3d. H 140. What is the subject of this ^ ? Why divide $1^50 by 8, tu gel the pri^e of 4 qts. of wheat? Why divide »'75 by 4 for the sanU purpose ? Give thi rule. ^ 1 141. PRACnCB. 185 KXAMPUBS FOR PRACTICS. 2. What costs 3 qts. of oil, at 8*94 per gal. ? Am. $*704. 3. What shall I receive for building 90 rods of road, at 11200 per mile ? Am. $337*50. 4. Bought 65 lbs. of pork, at $17'25 per barrel ; what did I pay ? J/ o -^ Z '> ' • Am, S5'60+. 5. What will 14 quires of paper cost, at $3'00 per ream ? /^ ^ % Am. $2' 10. 6. At Sf8'50 per month of 30 days for the rent of a house, what will be the rent for 18 days ? 2) $8*50 Solution.— Take half of the rent for ^ 30 days, which will be the rent for 15 days, 5) 4*25 /or 15 days, and one fifth of the rent for 15 days will be ^85 for 3 days, the rent for 3 days, and add together the rent for 15 and 3 days, the sum will be the $5*10 Am. ^^^ ^or 18 days. 7. What will a man's salary amount to in 7 months, at the rate of $500 a year ? 2) $500 6) 250=±for 6 TTumths. ^ 41*66+, 1 mo. Solution. — One half the year's salary will be the salary for 6 months, and one sixth of this the salary for 1 month. 291*66+, 7 mo. 8. What will be a man's salary for 8 months and 21 days, at $400 per annum, that is, by the year ? Am. $290. 9. What will 5 cord feet and 12 solid feet of wood cost, at $2'50 per cord ? Am. $1*80, nearly. * 10. What will 11 oz. of sugar cost, at 12 cents per pound? Am. $*082J. 11. What will 3| yards of broadcloth cost, at $4*00 pea yard ? Am. $14*50. % 14 !• To reduce shillings, pence, a7id farthings, to tht decimal of a pound, by impection, * There is a simple and concise method of reducing shillings, pence, and farthings to the decimal of a pound, by inspection. The reasoning in relation to it is as follows : 1^ of 20s. is 2s. ; therefore every 2s. is -j^, or *1£. Every •hilling is J^ == y^^, or *05£. Pence are roadily reduced to 16=** r-"^ 186 PRACTICE. T 141. farthings. Every farthing is -g^^. Had it so happened that 1000 farthings, instead of 960, had inade a pound, then every farthing would have been xtjVtti ^^ *001£. But 960 increased by gV part of itself is 1000 ; consequently, 24 far- things are exactly tS^tt' ^^ *025£., and 48 farthings are exactly i^^, or *050£. For, add ^V of any number of farthings to the number, and it will be reduced to thousandths of a poun^. If the farthings are 20, they will equal *020ff , which we will call *021, since f£^ is more than J of a thousandth. If the farthings are 14, they will equal *014U, which we will caU *015 for the same reason. But if the farthings are only 10, they will equal *010j^, which we call *010, since ^J is less than J a thousandth. If the farthings are 31 = *031J^ = 32^, we call them *032, for the same reason. And if the farthings be 42 = 42^ = 43^, we call them »044. The re- sult will always be nearer than ^ of 1 thousandth of a pound. Thus, 17s. 5Jd. is reduced to the decimal of a pound as fol- lows; 16s. = *8£. and ls. = *05£. Then 5|d. = 23 far- things, which, increased by 1, (the number being more than 12, but not exceeding 36,) is *024£., and the whole is *874£., the Ans, Wherefore, to reduce shillings, pence, and farthings to the decimal of a pound, by inspection, — Call every ttoo shillings one tenth of a pound; every odd shilling, five hundredthsT and the number of farthings^ in the given pence and farthings, so many thousandths, adding one, if the number be more than twelve and not exceeding thirty-six, and two, if the number be more than thirty-six. Note. — If the farthings be just 12 == *012^, they are equal to <0125 ; if 36, they are equal to '0375. 48 &rthiDgs ==: 12d. s=: Is. equal '05 of a pound, as above. fiXAMPUBS FOB PRACTICE. 1. Find, by inspection, the decimal expressions of 9s. 7d., and 12s. 0}d. A7is, *479£., and '603£. 2. Reduce to decimals, by inspection, the follow^ing sums, and find their amount, viz. : 15s. 3d. ; 8s. ll|d. ; lOs. 6Jd. ; Is. SJd. ; Jd., and 2Jd. Amount, £1'833. Questions. — 11141. What is the rule for reducing shillings, pence, and farthings, to the decimal of a pound, by inspection ? Whal IS the leasoning in relation to this rule ? •^tr-^ tl 1^, 143. PHICiafTAeE. 187 IT i4rft. ^0 re^fuce ^Ae decimal of a pound to sMltings^ pcnce^ and farthings^ by inspection. Reasoning as above, (tl 141,) the first three figures in any decimal of a pound may readily be reduced to shillings, pence, and farthings, by inspection. Double the first figure, or tenthsy for shillings, and if the second figure, or hundredths, he^e^ or more than five, reckon another shillhig ; then, after the five is deducted, call the figures in the second and third places so many farthings, abating one when they are above twelve, and two when above thirty-six, and the result will be the answer, within ^ a farthing. Thus, to find the value of *'&76£, by inspection : -r- *8 tenths of a pound ^=16 shillings. H)5 hundredths of a pound = 1 shilling. K)26 thousandths, abating 1, = 25 farthings, = Os. 6|d. *876 of a pound, = 17s. GJd. Ane. 1. Find, by inspection, the value of jff'523, and ,£'694. Ans. lOa. 5^., and Ids. 10^. ». Find the value of ^'47. Note. — When fhe decimal has but two figures, after taking out the shillings, the remainder, to be reduced to thousandthst wiU re- quire a cipher to be annexed to the right hand. Am. ds. 4|d. 3. Value the following decimals by inspection, and find then' ittiount, viz. : ^^'785, jg*357, ^'916, ^'74, ^'5, ^'25, j&K)9, and ^008. Ans. £3 12s. lid. PERCISNTAOE. H 143. 1. A man owns a farm of 320 acres, 5 per cent of which is marsh ; how many acres are marsh ? OPERATION. Solution. — Per cent, signifies hundredth part. g20 yThe nimiber placed before per cent, signifies how ^f^f, many hundredths are taken, being really the second ^•^ figure of a decimal fraction ; thus, 6 per cent, of 320 acres is *05 of that quantity, and since of implies 16'00 Qnestions.— IT 142. How may the first three figures of- any decimal of a pound be reduced to shillings, pence, and ferthrngs, by Inspection ? Explain the reasons for tiiis operation. 186 PERCENTAGE. % 143 multiplicatimi, we multiply 320 by <05, pointing off as in decimal fractions, and get the Ans. 16 acies. The finding of a certain per cent, or a certain number of hundredths of a quantity, is called percentage ; and it is per formed by the following Multiply the quantity by the rate per cent, written dec. mally as hundredths. Note. — Per cent, is horn the Latin, which signifies by the hi dred. 7 per cent, is *07. 25 per cent, is <25. 50 per cent is '50. 100 per cent is 1*00 (^%%, or the whole.) 125 per cent, is 1*25 (iH, more than the whole.) 1 per cent, is ....••• '0/ I per cent, is a half of 1 per cent., (| of 1 hundredth, or 10^00 of the whole,) .... *00^ per cent, is a fourth of 1 per cent, that is, ^ of -^ ^, =: '002^ per cent, is 3 times J per cent., that is, | of yj^, = *007f> i per cent., (| of a hundredth, that is, ^ of y^,) = '0012». 4^ per cent, is *04^ = *045, (the 5 expressing lOths of lOOths,) '045. BXAMPIiES. Write 2J per cent, as a decimal fraction. 2 per cent, is *02, and ^ per cent, is *005. Ans. '025. Write 4 per cent, as a decimal fraction. 4J per cent 4| per cent 5 per cent 7J per cent. 8 per cent. 8f per cent. 9 per cent. 9J per cent. 10 per cent 10 J per cent 12 J per cent. 121 per cent — 133^ per cent. BXAMPIiES FOR PRACTICE. 2. A farmer gives 10 per cent, of 460 bushels of wheat for threshing ; how many bushels does he give ? Ans, 46 bushels. 3. A farmer rented ground on which 409 bushels of oats were raised, receiving 30 per cent, for the rent ; how many bushels did he receive ? Ans, 122*7 bushels. ■ ' ' — Qnestions* — ^ 143* What does per cent, signify ? — percentage ? — the number before per cent. ? How is *05 of a quantity obtained, and why ? Give the rule for percentage. How is any per cent, from 1 to 99 expressed? 100 per cent. 125 per cent. ? j| per cent. ? 4j| per cent f 1 T 143. raRCENTAGB. 18B 4. A beef weighs 895 lbs., of which 9 per cent, is bone ; what does the meat weigh ? Ang. 814*45 lbs. 5. A schooner, freighted with 725 barrels of flour, encoun- tered a storm, when it was found necessary to throw 28 per cent of the cargo overboard ; how many barrels were thrown oyerboard, and how many were saved ? Afu. to the last, 522 barrels. 6. A forwarding merchant agreed to transport 2000 bush- els of corn, worth $692*75, from Bufialo to Albany for 12J per cent, on its value ; what was the cost of transportation ? Am. t86*59|. 7. A farmer had a flock of 639 sheep, which increased 33| per cent, in 1 year ; how many sheep had he at the expira- tion of the year ? Ans. 852 sheep. 8. A man owincf a debt of $1942*711, pays 16§ per cent, of it ; how much of it remains due ? Ans, $1624*595 -|-. 9. A man, worth $4861, lost 284 per cent, of it by en- dorsing with his neighbor ; how mucn of it did he lose ? Ans. $1385*385. 10. What is } per cent, of $115? Ans. $*862J. 11. What is I per cent, of $376 ? Ans. $3*29. 12. A gentleman, worth $4280, spent 15^ per cent, of his property in educating his son ; how much aid the son's edu- cation cost his father ? Ans. $663*40. 13. A merchant has outstanding accounts to the amount of $1960 ; 22 per cent, of which is due in 3 months, and the remainder in 6 months. What is the amount due in 3 months ? in 6 months ? Ans. to the last, $1528*80. 14. A merchant who fails in business pays 63 per cent, on his debts; what does a man receive whose demands are $2465 ? ' Ans. $1552*95. 15. What does another man lose, whose demands are $3615 against the same merchant ? Ans. $1337*55. 16. A young man is left with $5000, and loses 15 per cent, in paying too high a price for a farm, 15 per cent, of the remainder in selling the farm for less than its value ; he ex- pends 15 per cent, of what is left in an excursion to the west, 15 per cent, of what he has when he gets back in an unfortu- nate investment in railroad stocks, and 15 per cent, of the resi- due in trade ; what has he then left ? Ans. $2218*526 [-. Note. — Under the general subject of Percentage will be consid- ered Insurance, Stocks, Brokerage, Profit and L^, Interest, Dis- count, Ck>mmis8iQi, Bankruptcy, Partnership, Banking, Taxes and Duties. 1^ VWf^mTAOSL % 14f Insurauce. ^ 144* Insurance is security to individuals against Ipsp of property from fire, storms at sea, &c. Companies incorporated for the purpose, having a certain capital to secure thei,r responsibility, insure property at so much per cent, a year. When any property insured is de- stroyed by the agent insured agamst, the company pays to the owner the sum for which it is insured. The sums paid by the several individuals insured, make up the losses, and pay the company for doing the business. Premium is the sum paid for insuraacQ. Policy is the writing of agreement. An Underwriter is an insurer, whether it be an incor- porated company or an individual. Insurance at sea, called Marine insurance, is usually for a certain voyage. It is sometimes effected by an individual ; it is then called out-door insurance. The rate per cent, of insurance is in proportion to the risk. Property is not insured for its entire value, lest it should be fraudulently destroyed* BXAMPXiES FOR PRACTICB. 1. What is the annual insurance of $1000 on an academy, at i per cent ? Ans, $5. 2. Insured $14500 on a factory at 1| per cent, per annum; what was the premium ? Ans, $253'75. 3. What is the premium for insuring $6000 on a store and goods at I per cent. ? Solution. — At 1 per cent, the sum is as many cents as there axe dollars, or 6000 cents, which reduced is $ 60^00, and | per cent, is i of this, or, Ans. $ 45. Note. — In this manner the percentage on any sum at 1 per cent, or less may be calculated with ease. ^ 4. What must be paid for insuring $800 on a farm houses at J per cent. ? Ans. $2. ■ III — -ii^— ^^—i^M I , I ^— i^ Questions. — % 14A. What is insurance? Who insure? How is insurance estimated? Who pays, if the property be destroyed? How much? What remunerates the company? Wh^is premium? — policy ? — an underwriter ? — marine insurance ? ^ow is marine in surance often effected? What is it then called? What is life insurance, and for whatpurpose is it effected ? What is said of health insurance I Example. Why is not property insured for its entire value ? T 145. FBRCENTAOa 191 5. The Marine Insurance Coinpany insures $17500 on the cai^o of the ship Minerva, from Boston to Constantinople, at 2 per cent ; what is the premium ? Ans. $350. 6. Amos Lawrence insures $34000 on the ship Washing- ton and cargo from Canton to Boston, at 8^ per cent ; what does he receive ? Ang. $2805. 7. The New England Life Insurance Company insures $2000 on a person's bfe for one year at a premium of IW^ per cent. ; what is the sum ? Am. $21. Note 1. — Life insurance is efl^cted that the heirs of the individaal, in case of his death, may receive the sum on which the premium is paid. The insurance is usually for one year, for seven years, or for life, and the annual rate per cent, is determined by a careful estimate made from bills of mortality of the probable chances of death with per- sons of different ages. Note 3.-^ The premiums of health insurance companies, which have lately been organized, are specified suras to be paid annually in proportion to what is received weekly in case of sickness. Thus in the Massachusetts Company, $ 5^00 a year is paid by a person at the age of 30, to secure $4'00 a week in sickness. I'he premium it de- termined from a careful estimate of the probabilities of health. Mutual InQurance. V 14tS* The rate at which companies can afibrd to in.<ure is estimated from the probable losses that will occur. But when the losses are small, large profits are made by the com- pany. Or the losses at some time may be greater than the means at the command of the company, whereby its capital will be annihilated, while the losses of the insured will not be fully made up. Hence, mutual insurance companies have been formed, to average the losses that may actually occur, which it is the aim of all insurance to do. Each one gives a premium note of so much per cent, on the property which he wishes to insure, the rate being determined by the risk of the property. The amount of these notes are the capital of the company, and a per cent, is paid down on them^ to furnish money for Qnestions. — If 145* How is the rate of insurance in ordinary companies estimated? What objections to this method? What is the aim of £^ insurance ? Describe the premium note. How is the rate deteiw mined? How is money procured for use? What is the capital of the company? Why, and on what, are assessments made ? Appl^ thi^ principles to Ex. 1, and its solution. ^ l^ PERCENTAGE. T146 nnmediate use. Any losses that occur more than tks are averaged on the premium notes. BXAMPIiES FOR PRACTICE:. 1. What sum is paid by a farmer for insuring $1500 on his buildings for five years in the Cheshire Mutual Insurance Company, the premium note being 7 per cent., of which 3 per cent, is paid down, and assessments paid afterwards of 2, 1J» Sj, and I per cent. ? OPERATION. $1500 *07 105'00 Solution. — First find the amount of the premium »10i note, which is 7 per cent, of $ 1500 = $ 105, and 3 + 1 2-hl4 + 31+i = 104 POT cent, of $ 105=11*02^, Ans, 52J 10 50 .$ll*02i 2. What sum must be paid for insuring $2845 on a store for the same time, and with the «ame assessments, the pro mium note being 12 per cent. ? Ans, $35'847. 3. What ijfiust be paid as above, premium note 15 per cent. ? Ans. $44*808f . 4. Insured $5000 on a flouring mill for five years, in the Tompkins Co. Mutual, premium note 22 per cent, of which 4i per cent, was paid down, and 2J per cent, in assessments ; what did it cost per year* Ans, $15*40. 5. Insureck for five years $32P0 on a house of worship in the Vt. Mutual, premium note II per cent., on which 8 J per cent, was paid down, and four assessments were made re- spectively of IJ, 2 J, 2, and f per cent. ; what is the whole sum paid ? Ans, $35*49 J. 6. How much more would it cost to insure the samS prop- erty for the same time in the -^tna Insurance Company at J per cent, each year? Ans. $44'50|. 7. What must be paid annually to insure $750 for five years on a library, premium note 6 per cent., paid down 4 ^per cent., sum of assessments 9 J per cent. ? Atis, $1*21 J. 8. Insured for five years $900 on furniture, premium note 5 per cent., sum of payments on it 6 per cent. ; how much is paid : ^ Ans, $2*70. ^ "I T 146. PSaCENTAOB. 193 ^ stocks. V 14W. In the construction of a railroad, which costs Bay $200000, the sum is divided into shares usually, of $100, each individual paying ihe amount of a certain number of shares, which are called his stock in the road. The one thus paying towards the road is called a stockholder, and is remu- nerated by a proportional share of the profits. It follows, of course, that the road may be so profitable that each share will be worth more than $100 ; the stock is then said to be above par. If the road is unprofitable, a share will be worth less than $100, and the stock is said to be below par. When a " share is worth just $100, the stock is said to be at par. The stockholders together constitute the railroad company, and the sum of the shares is the capital of the company. Manufactories, too large for individual enterprise, banks, kc.y are conducted in a similar manner. When governments borrow money, the sum each lends is said to be his stock in what are called the government funde. BXAMPIiES FOR PRACTICE. 1. What is the value of 35 shares in the Fitchbuig rail- road, at 120 per cent. ? i%% of $3500 = how- much ? Ans. $4200. 2. Sold 15 shares of the Eastern railroad at 7 J per cent, advance ; what sum did I receive ? Ans, $1612*50. 3. What do I pay for 20 shares in the Old Colony rail- road, at 1 J per cent, below par ? Ans, $1975. 4. What are 28 shares in the Vt. Central railroad worth, It Hi per cent, below par? Am. $2485. 5. wliat are 45 shares in the iBxchange Bank worth, at 8 per cent, below par ? Am. $4140. 6. What are 30 shares in the Western railroad worth, at 9J per cent, above par ? Atis. $3285. 7. For what must I sell $5000 U. S. 6 per cent, stock, that is, stock on which 6 per cent, per annum i^ to be paid, at 1 J per cent, discount ? ' Am. $4925. 8. What is $3200 in the Amoskeag Cotton Manufacturing Co. worth, at 17 per cent, above par ? Ans. $3744. l^nestions. — IT 1^« How is a railroad built ? What is a share ? — a stockholder, and how remunerated ? When, and why, is stock above par? below par ? — at par ? What is the company ? - he capital ? What else are established on the same plan ? What ar# -ovemment funds? 17 194 PERCFNTAGE. 1[ 147, 14a 9. What is $2000 in the Ocean Steam Navigation Co. worth, at 2 per cent, advance ? Ans. ^040. 10. Bought 9 shares in the Western Transportation Co., at 4 per cent, below par ; what did I pay ? $864. Brokerage. ^T 1^7 • Brokerage is an allowance made to a dealer in money, stocks, &c., who is called a broker. The allowance is generally a certain per cent, of the money paid out or re- ceived. EXAMPLES FOR PRACTICED 1. A western merchant procures $8500 in bills on N. E. banks of a Boston broker, for Ohio money, the broker charg- ing 2 per cent, for the accommodation ; what brokerage does he pay ? Am $70. 2. A drover exchanges $2240 of country money for city bills, paying i per cent, on his country money ; what does he receive? ' Ans, $2237*20. 3. A broker is directed to buy 150 shares of the N. Y. and Erie railroad stock. He pays $92 per share, and re- ceives I per cent, on the money advanced ; what does he receive on the whole ? Am. $103'50. 4. What must I pay a New York broker for $5000 of city hank bills, in bills on eastern banks, at J per cent. ? Am. $5012'50. 5. A broker sells for an individual 90 shares of the Fitch- burg railroad for $125 a share, receiving 1 p^r cent, on what money he gets; what does he receive ? Am. $112^50. 6. Bought $6000 in gold coin, paying the broker 1 per cent, for it ; what does he receive ? A71S. $60. 7. Sold $5200 in gold sovereigns, at a discount of J per cent., for good bank bills, which are more convenient for me to carry ; how much in bills do I receive ? Am. $5174. Profit and Loss. IT 148. 1. Bought cloth at 40 cents a yard ; how must I sell it to gain 25 per cent. ? Questions. — ^ 14T. What is brokerage ? — a broker? How is orabeiage calculated? i 1 149. PERCEI^IAGI!^ 196 SeLUTi«>ifl. — When the price at which g<>od8 are bought is giren to find the price for which they must be sold, in order to gain or lose a certain per cent., the calculation is by the general rule for percentage. Ans. $'50. Note. — The profit or loss must be added to or subtracted from the price of purdiase. BXAMPI^BS FOR PRACTICB. 2. Bought a hogshead of molasses for 960 ; for how much must I sell it to gain 20 per cent. ? Atu* S72. 3. Bought broadcloth at S2,50 per 3rard; but, it being damaged, 1 am willing to sell it so as to lose 12 per cent. ; how much will it be per yard ? Ans. $2*20. 4. Bought calico at 20 cents per yard ; how must I sell it to gain 6 per cent. ? 10 per cent. ? 15 per cent. ? to lose 20 per cent. ? Ans. to the last, 16 cents, per yard. Interest. V 149. Interest is an allowance made by a debtor to a creditor for the use of money. JPer anvum signifies /or a year. The rate per cent, per annum is the number of dollars paid for the use of 100 dollars, or the number of cents for the use of 100 cents for 1 year. Note. — Percentage has hitherto been computed at a certain per cent., usually without regard to time; interest is computed at a certain per cent, for one year, and in the same proportion for a longer or shorter time. Principal is the money due, for which interest is paid. Amount is the sum of the principal and interest. Legal interest is the rate per cent, established by law. Usury is any rate per cent, higher than the legal rate. The legal rate per cent, varies in different countries, and in the different States. It is Questions. — If 148. What do y;on understand by profit ? — by loss ? What are given, and what required, in this ^ ? How found ? 1F140. What is interest? How is it computed? What- does par annum signify ? — rate per cent, per annum ? What distinction do you make between percentage and interest ? What is the principal ? — the amount' — legal interest ? — usury ? What is the legal rate per cent, in each of the states? When no rate per cent, is mentioned, what rate per cent, is understood ? How is interest for one year computed ? — for more than a year ? 196 PEKCfiNTAOSL IT IM 6 per cent: in all the New England States, in Pentt&yfVa- nia Delaware, Maryland, Virginia, N. Carolina, Tennessee; Kentucky, Ohio, Indiana, Illinois, Missouri, Atkansas, New Jersey, District of Columbia, and on debts or judgments in favor of the XJ. Stettes. 7 per cent, in New York, S. Carolina, Hichigsan, Wiscon- sin, and Iowa. 8 per cent, in G>eorgia, Alabama^ Florida, Texds, and Mis- sissippi. 5 per cent, in Lotiisiana. When no rate pes cent* is named, the legal rate per cent, of the stele where the business is transacted, is always under- stood. At 6 per cent., a sum equal to f^ of l^e ];»rincipal lent or due is paid for the use of it one year ; at 7 per cenl., a sum equal to t5tp<*^ i*» ^^'^ so of any other rate per cent. Hence, To find the interest of any sum for 1 year, is to take such a fractional part of the principal as is indicated by the rate per cent., as in percentage, and by the same rule ; that is, we multiply the principal by the rate per cent. For rnore years than 1, multiply the interest for 1 year by the number of years. Interest is computed not only for one or more years,, but in " the same proportion " for months and days. T 150. To find the interest on any sum for months at any rate per cent. 1. What is the interest of $216*80, at 7 per cent., for 1 month ? for 2 months ? 3 mo. ? 4 mo. ? 6 mo. ? 6 mo. ? 7 mo. ? 8 mo. ? 9 mo. ? 10 mo.? 11 mo.? First, find the interest for 1 year, and then for the months take fractional parts of the interest for 1 year. NoTB. — The pupil will readily perceive OPERATION. methods of simplifying and shortening the $216'80 Principal, operation, according to If 140. Thus, he *07 rate per ct. may take i of the interest for 1 year, that ^ is, the int. for 6 months, and ^ the int. for $15*1760 Int. 12 mo. ^ months, = the interest for 2 months, and add these together for the interest 8 montlis. Qaestions* — If 150* How is interest calculated for months? What part or parts do we take for the interest 3 months ? — 4 months ? — 5 months? — 6 months? — 7 months ? — 8 months ? — 9 months f 915a760 IfU.l2fno. For 1 mo. take ^ of tiie int. for J2 mo. = $1*264 + Am. 2*529 + " 2 mo. u " 3 mo. (( " 4 mo. (( " 5 mo. C( " 6 mo. (( " 7 mo. C( " 8 mo. it " 9 mo. M " 10 mo. u "11 mo. (( « = 2*529-^ «* " == 3*794 «« « = 5*058+ « " = 6*3234-. " " = 7«588 " --iV " = 8*852+ " «• =10*117+ « «• s=: 11*382 " = 12*646+ = 13*911+ 4< it BXAMPLSJS FOR PAACTICB. 2. What is the interest of $450 for 9 months, at the rate in Louisiana ? What is the amount ? Ans. to the last, $466*87 J. NoTB. — The amount, which is the sum due, is found bj adding the principal and interest together. 3. What is the amount of $87*50 on interest 7 months, at the rate in Georgia ? '- Ans. $91*583. 4. What will be the interest of $163, for 4 months, at the rate in S. Carolina ? Am. $3*803. 5. What will be the interest of $850, for 10 months, at th« rate in Kentucky ? Ans. $42^50. IT 1<5| • To find the wjterm on miy stum for days, 1. What is the interest of $216*80, at 7 per cent., for 1 day ? for 2 days ? for 3 daya, and so on, to 29 days ? • Note 1. — In computing interest, 1 monlh is reckoned 30 days. First, find the interest for 1 year, then for 1 month, as in Ex. 1, last IF. These operations we need not repeat here, but take the interest as there found, $1*264 for 1 month, of which we may take parts, thus : Questions* — IF 1 5 1 . How many days are called a month m com- puting interest ? How is the interest for &ys found ? -^ for 3 days ? — 5 days? — 8 days? — 10 days? — 12 days? — 18 days? — 21 days? — 25 days ? — 28 days ? Explain the jainciple of the direction how to take 3\) of a number. What is the amount, and how found? 17# 198 PERCENTAGE. HlBSt. For 1 day take ^ of Int. for 1 mo. = $*042 - - C( <t 5 days 6 " 10 " 15 i i i u ti a a OPERATIOM 3)$r264 Int. 1 tno. 1 day. 5 days 6 *210 - - *252-- *421-- *632 *758 *842 (( « (( (i 10 « 15 (I (( i( (i "18 « " 3 times-t,orJ+i\y== *758 "18 " 20 " " 2 times i = *842 « 20 " NoTK 2. — To get 7j\j- of a number, divide it by 3, set^g the l^t figure of the quotient 1 place towards the right, as in the opera- tion. For 2 da3rs, take 2 times the interest for 1 day. For 9 days, J of int. for 30 days plus ^ of | of int. for 30 days. For 11 days, \ plus ^. For 16 days, take 4 times ti^e interest for 4 days, or ^ the int. for 1 month, -}- the int. for 1 day.' For 20 days, we may take 2 tames the int. for | of 1 month. For 25 days, 1 4" i ^^^ ^°^' ^^' ^ month, varying a^r this manner as may suit our convenience. KXAMPUBS FOR PRACTICK. 2. What is the interest of $400, at the rate in Alabama, for 9 days ? Am. $*80. 3. What is the interest of 875, for 19 days, at the rate in Florida? - Ans. $'31%. 4. What is the interest of $500, for 25 days, at the rate in Texas ? Am. $2*777. IT ISSt* WAen the time is expressed in mcrre than one de* nomination^ as in days and rrumthsy or years^ months^ and days. 1. What is the interest of $32*25, for 1 year 7 months 19 days, at ^ per cent. ? OPERATION. $32*25 Principal. '04 5 rate per cent. 16125 12900 $1*45125 Int. for 1 year. i Bt. of 12 mo. = *7256 - - I « 6 " =4209-- I " 30 da. = *0604-- l " 15 " =*0120-- I u 3 « — «0040-- 6 TTumths. 1 monthr- 15 days. 3 " 1 daiy. Am. $2*3741 -f- Int. forlyr.7 mo. 19 da. i i 1 152. PERCENT AGB. 199 Hence, To find he interest on any tum^ for any time, at any rate per cent., this I. For 1 year. Multiply the principal by the rate per cent., pointing off as in decimal fractions, and the product will be the interest for 1 year. II. For more years than 1, multiply the interest for 1 year by the number of years. * ni. For months. First find the interest for 1 year, of which take such fractional part as is denoted by the given number of months. IV. For days. Take such part of the interest for 1 month as is denoted by the given number of days. V. For years, months and days, or for any two of these de- funninations of time. Find the interest of each separately, and add the results together. EXAMPIiBS FOR PRACTICC 1. What is the interest of $84, for 1 year 9 months 20 days, at the legal rate in Alabama? Ans, $12*133. \ 2. What is the interest of $147, for 2 years 8 months 12 days, at the rate in Michigan ? Ans, $27*783. \ 3. What is the interest of $248, for 2 years 6 months 20 days, at 9 per cent. ? Ans. $57'04. 4 What is the interest of $161*08, for 11 months 19 days, at the rate in N. Y. ? Ans. $10*931 -f-. 5. What is the interest of $73*25, for 1 year 9 months 12 days, at 8 per cent. ? Ans, $10*45 -}-. 6. What is the interest of $910*50, for 3 years 9 months 26 dtays, at 7 per cent. ? Ans. $243*609 -f-. 7. What is the amount of $185*26, in 2 years 3 months 11 days, at 7| per cent. ? An^. $216*947 -f*. 8. What is the interest of $656 from Jan. 9 to Oct. 9 fol- lowing, at J per cent. ? <ttfi«5 D • • 7 Solution.— Remove ^P^ -rnnctpac, the separatrix two places 2^6*56 — Int IvT at\ ner cpnt ^ *^® ^^^' *"^ ^^® ^^™ 4, ; ooD _ mi, lyr.ati per cent, -^^y^^-^i ^^^^^^ the in- 2)3*28= ** ** J ** terest for 1 year at 1 per cent., T[ 149, ^ of which 2) 1*64= ** 6 mo, ** ** - wUl be the interest for 1 *82 = *» 3 TJio, ** " year at i per cent , of A ^rt A^ r /^ which take fractionai Am, $2*46 Int, 9 mo, at ^ per cent, parts for 9 months. Qncstions. — ^ 152. Explain the operation, Ex. 1. What is the general rule? 200 PERCENTAGE. IT 1S3. 9. What is the interest of $46*28, for 2 years 3 months 23 days, at 6 per cent. ? Am. $5*354 -f . 10. What will he the amount of $175*25, in 5 years 8 months and 21 days, at 6 per cent. ? Am. i235'448 +. 11. What wiU be the amount of $96*50, for 1 year 11 months 29 days, at 12j per cent. ? Am. $12(^691 +. Note. — Aul2h per cent., | of the principal will be the interest for 1 year. 12. What is the interest of $54*81, for 1 year and 6 . months, at the rate in Louisiana? Atis. $4*11. 13. What is the interest of $500, for 9 months 9 days, at the rate in Georgia? Ans. $31. 14. What is the interest of $62*12, for I month 20 days, at 4 per cent. ? Am. $*345. 15. What is the interest of $85, for 10 months 15 days,ut 12J per cent. ? Am. $9*296.'^" 16. What is the amount of $53, at 10 per cent., for 7 months ? Am. $56*091. 17. What is the interest of $327*825, at the rate in Flor- ida, for 1 year ? Am. $26*226. 18. What is the interest of $325, for 3 years, at the rate in Pennsylvania ? Am. $58'50. 19. What is the interest of $187*25, for 1 year 4 months^ at the rate in Delaware ? Am. $14*98. 20. What is the interest of $694*84, for 9 months, at 10 percent.? Am. $52*113. 21. What is the interest of $32*15, for 1 year, at ^ per cent. ? Am. $1*446 -j-. 22. What is the amount of $60Q, in 2 years, at the rate in New England ? Am. $672. 23. What is the interest of $57*78, for 1 year 4 months 17 days, at 4 per cent. ? • Am. $3*19. 24. What is the amount of $298*59, from May 19th, 1847, , till Aug. 11th, 1848, at the rate in Texas? (See 1[ 127.) . / 25. What is the amount of $196, from June 14, 1847, till April 29, 1848, at 5| per cent. ? ' Am. $205*861 +. To compute interest for rrumths and days^ when the rate is 6 per cent. L Of One Dollar. IT tSS* A method brought out in the ** Scholar's Arith- metic, ' 1801, and revised in ** Adams* New Arithmetic," T153 PBROENTAGS. 901 1827, vill be preferred by many to the ' foregoing, in ^ose states ^here the legal rate is 6 per cent. induction. The interest on $1 for 1 year, lit 6 per oent, oeing ♦06 cents, is •01 cent for 2 months, *005 mills (or j- a cent) for 1 month of 30 days, and •001 mill for every 6 days ; 6 being contained 6 times in 30i Hence, it is very easy to find, by iTUpection, the interest of 1 dollar, ^t 6 per cent., for any given time. The cents will be equal to half the greatest even number of months. The mills will be 5 for the odd month, if there be one, and 1 for every time 6 is contained in the given number of days, with such Part of 1 mm, as the days less than 6, are part of 6 days. 1. What \s the interest of 81, at 6 per cent., for 9 months 18 (lays ? Solution. — The greatest even namber of months is 8, the interest for which will be $^04 ; tiie mills, reckoning 5 for the odd month, and 3 ftur the 18 (3 tiiilBS 6 = 18) days, will be $'008, which, added to the cents, give 4 cents 8 mills for the interest of $1 for 9 months and 18 days. Ans, $*048. 2. What will be the interest of $1 for 6 months 6 days ? 6 months 12 days ? 7 months ? 8 months 24 days ? 9 months 12 days ? 10 months ? 11 months 6 days ? 12 months 18 days ? 15 months 6 days ? 16 months ? \ 3, WTjat is the interest of $1 for 13 months 16 days ? Solution. — The cents will be 6, and the mills 5, for the odd month, and 2 for 2 times 6 s= 12 days, and there is a remainder of 4 days, the interest for which will be such part of 1 mill as 4 days is paart of 6 days, that is, f = | of a mill. Ans, $'067 1. 4. What is the interest of $1 for 12 months 3 days ? ^ I ^- I I J I IL 1 . II I I ■ ■! H I— Questions. — 1 153* At 6 per cent., what is the interest of $1 for 1 year? — for 2 months? — for 1 month of 30 days? — for every 6 days? How, then, may the interest of $1, at 6 per cent., foraijy given time, be foand by inspection? If there is no odd mcmth, aad the num- ber of days be less than 6, what is to be done ? Why ? The interest on $1 for a number of days less than 6, is what ? How do you fiad it vrn^ ren m the ep^ami^es which have been given ? 202 PERCENTAGE. iri54. NoTB. — * If there is no odd month, and the nuambet of days be less than 6, so that there are no mills , a cipher must be put in the place of mills ; thus, for 12 months 3 days, the cents will be '06, the mills 0, the 3 days k a mill. Ans. $<06^. 5. What will be the interest of $1, for 2 months 1 day ? 4 months 2 da^ s ? 6 months 3 days ? 8 months 4 days ? 10 months 6 days ? for 3 days ? for 1 day? for 2 days ? for 4 days ? for 6 days ? IT 154. 12 days ? il^zf . to the last, $'000f . II. Of any- Sum. 1. What is the interest of $75, for 10 months FIRST OPERATION. $<052 Int. on $1. 75 260 364 83*900 Int, of $76. SECOND OPERATION. $75 *052 Solution. — We find the interest of $1, by the last ^f, which is $'058, and 75 times this sum, as in the first operation, will be the interest of $75 ; or, since either factor may be made the multiplicand, (^[21,) wo multiply $75 by *052, thus taking 52 thou- sandths of the principal, for that is the part taken, when the if^terest of each dollar is $*052. 150 375 $3*900 Int. of $75. 2. What is the interest of $56^13, for 8 months 5 days ? Solution.— We find the interest of $1 for the time to be $*040f , and we multiply the principal by *040|^. As 1 thousandth of the multipli* cand is taken (\ mill or thousandth being the unit of the multiplier) for every 6 days, for the days less than 6, we take such fractional part of the mul OPERATION. 3 12) 51^6*13 Pnncipal. *040f 224520 Int.forSTno. fmultipi =2806 « 3 days. " =1871 " 2 «« $2*29197 Int. for Smo.5 days. Am. ^ N T 154. FEBCENTAGE. 203 tiplicand is the odd day or days is of 6. Thus, for 3 days, we take i| of the multiplicand, which will he the interest for that time in mills. For 2 days, we take ^ of the multipUcand. Adding together the in- terest for 8 months, 3 days, and 2 days, the sum wul be the interest for 8 months and 5 days. Note 1. — As the interest of $1 for 6 days ia 1 mill, that of $10 for the same time will be 10 mills = 1 cent. Hence, if the sum on which interest is to be cast be less than $10, the interest, for any num- ber of days less than 6, will be less than 1 cent ; consequently, in busi- ness transactions, if the sum be less than $10, such days need not be regarded. Hence, To find the interest of any given sum^ in Federal Money y for any length of tvme, ut 6 'per cent,, RUL.E. I. Pind the interest on $1 for the given time by inspec- tion. II. Multiply the principal by this sum, written as a deci- mal, and point off the result as in multiplication of decimals. ESXAMPLiES FOR PRACTICIJ. 3. What is the interest of $194, for 4 months 12 days? Ans. $4'268. 4. Interest of $263'48, for 2 mo. 21 days ? • $3'556. 5. Amount of $985, for 5 years 8 months ? $1319*90. 6. Interest of $87' 19, for 1 year 3 months ? $6*539. 7. " of $116*08, for 11 mo. 19 days ? $6*751. 8. ** of $200, for 8 mo. 4 days ? fS'133. 9. " of $0'85, for 19 mo. ? $'08. 10. " of $8*50, for 1 year 9 mo. 12 days ? $*909. 11. « of $675, for 1 mo. 21 days? $5'737. 12. " of $8673, for 10 days ? $14*455. 13. " of $073, for 10 mo. ? _ $'036. 14. " of $126'46, for 9 mo. ? \ $5*69. 15. ** of $318*, for 10 mo. 16 4ay»1 $16*748. 16. ** of $418', for 1 year 7 mo. 17 days ? $40*894. 17. " of $268*44, for 3 yrs. 5 mo. 26 ds. ? $56*193. 18. " of $658, from Jan. 9 to Oct. 9 fol- lowing? $29*61. Qnestions, — If 154. After the interest of $1 is found, how is the mterest of $75 found, by the first operation, Ex. 1 ? — by the second operation ? "Why ? In what denomination is the |- cf the multiplicand *taken in Ex. 2, and why ? What is said of the interest of $10 for less than 6 days, and Way? Give the rule. How may the mterest of ani sum be found for 6. C ays ? — for less than 6 days ? i^P^^"^W*^^^™p^^^^^P"«*^»^^^^^^w»^""^~"^"^»»"^w^^^»"^^"»^^^^^^»^"""-«»' J20A PERCENTAGE. f 156. 19. Interest of $96, for 3 days ? Note 2. — The interest of $1 for 6 days being 1 mill, the doHarB themselves express the interest in mills for sis days, of which we may take parts. 20. ' Interest of $73*50, for 2 days ? 21. " of $180'75, for 5 days ? 22. " of $15000, for 1 day ? IT 155. When 6 per cent, interest is required for a larg-e QUQ^er of years, it will be more convenient to find the inter- est for one year, and multiply it by the number of years ; after which, find the interest for the months and days, if any, as usual. 1. What is the interest of $520'04, for 30 years and 6 months ? ^ OPERATION. $520-04 Principal. »06 2) $31*2024 Int. 1 year, 30 $936*0720 " SO years, $ 15*6012 ** 6 mo. $951*6732 " 30 years. 6 mo, Ans, $961*6r3. 2. Whjat is the interest of $1000, for 120 years ? Ans. $7200. 3. What is the^terest on $400 for 10 years 3 months and 6 days ? ^ Ans. $246*40. 4. What is the interest of $220, for 5 years ? for 12 years ? 50 years ? Ans, to the last, $660. 5. What is the amount of $86, at interest 7 years ? Ans, $122*12. 6. What is the amount of $750, on interest 9 years 4 mo. 14 days? ibw. $1171*75. l^uestioiis. — T \56. How do we get 6 per cent, interest for a large number o^ years ? H >w, when there are also months and dajrs f ^ 156, 157. PERCENTAGE. 206 To find the interett on pounds^ skUUngSt and pence. f 196. 1. What is the interest of £36 9s. 6|d., for 1 year, at 6 per cent. ? Reduce the shillrngs, pence, &c.) to the decimal of a pound, by in- Section, (^ 141,) then proceed in all respects ad in federal money avin^ found the interest, reverse the operation, and reduce the first three decimals to shillings, &c., by inspection, (^ 142.) Ans. £2 3s. Od. 2. Interest of £36 10s., for 18 mo. 20 days, at 6 per cent. ? Ans, £3 8s. l^d. 3. Wterest of £95, for 9 mo. ? Ans. £4 5s. 6d. 4. What is the amount of £18 12s., at 6 per cent, interest, for 10 months 3 days ? Ans, £19 10s. 9Jd. 5. What is the an^ount of £100, for 8 years, at 6 per cent. ? Ans, £148. 6. What is the amount of £400 10s., for 18 months, at 6 per cent. ? Ans, £436 10s. lOd. 3qr. 7. What is the amount of £640 8s., at interest for 1 year, at 6 per cent. ? for 2 years 6 months ? for 1 years ? Ans. to the last, £1024 12s. 9Jd. 8. What is the amount of £391 17s., for 3 years 3 mo., at 4^ per cent. ? 9. What is the amount of £235 3s. 9d., from March 5, 1846, till Nov. 23, 1846, at 5J per cent. ? Am, £2U 8|d. To calculate interest on notes , ^c, tohen partial payments have been Tnade. IT IS7» Payment of part of a note or other obligation, is called a partial payment. It has been settled in the Supreme Court of the U* States, and their practice adopted by nearly all the states in the Union, that payments shall be applied to keep down the in- terest, and that neither interest nor payment shall ever draw interest. Hence, if the payment at any time exceed the inter- est computed to the same time, that excess is taken from the principal; but if the payment be less than the interest, the principal remains unaltered. Hence, the Questions*— ^ 156* How do we proceed, When the principal u pounds, shillings, and pence ? 18 * •• ■ J« '•- — >>— *.^. — ** V 206 PERCENTAGE. H 157. RUUS* Compute the interest on the principal to the time when the payment, or payments, (if the first be less than the interest,) shall equal or exceed the interest due ; subtract the inteaest Irom the payment, or mm of the payments, made within the time for which interest was computed, and deduct the excess from the principal. The remainder will form a new principal, with which pro- ceed as with the first. 1. S116'666. Boston, May 1st, 1842. For value received, I promise to pay James Conant, or order, one hundred and sixteen dollars sixty-six cents and six mills, on demand, with interest. Samuel Rood. On this note were the following endorsements : Dec. 25, 1842, received $16*666" July 10,1843, »' $ 1*666. Note. — In finding the ft*m«s Sept. 1, 1844, " $ 5*000 > for computing the interest, c<Mir June 14,1845, *« $33*333 suit 1 127. AprU 15, 184i, ** $62*000 J What was due August 3, 1847 ? A7is. $23*775. Note 1. — The transaction being ifi Massachusetts, the rate of in- terest will be 6 per cent. The first principal on interest, from May 1, 1842, $116*666 Payn>^^ Dec. 25, J842, (exceeding in- teresTdue,) $16*666 Interest to time of 1st payment, . . 4'549 12*117 Remainder for a new principal, .. . . $104*549 Payment, July 10, 1843, less than inter- est then due, $1*666 Payment, 'Sept. 1, 1844, less than inter- est then due, $5*000 Amount carried forward, $6*666 Questions* — TJ 157. What is a pajtial payment of a note or obli- gation ? What court has established a rule for computing interest on notes, &c., on which partial payments have been made ? How exten- sively is this rule adopted? Repeat the rule. What is -the great funda- mental principle on which this rule is based ? What is customary when notes with endorseme its are paid within one year of the time they are given ? A V % 157. PERCENTAGE. 207 Amount brought forward, t6*666 Payment, June 14, 1845, . . . $33'333 Amount, (exceeding interest due,) $39*999 Interest from Dec. 25, 1842, to June 14, 1845, (29 mo 19 lays,) . . . 15*490 Payment, April 15, 1846, (exceeding in* terestdue,) $62*000 Interest from June 14, 1845, to April 15, 1846, (10 mo. 1 day,) . . . 4*015 24*509 $80^040 57*985 Remainder for a new principal, . . . $22*055 Interest due August 3, 1847, from April 15, 1846, (15 months 18 days,) . . . . ' 1*720 Balance due Aug. 3, 1847. . . Ans, $23*775 2. $867*33. Buffalo, Dec. 8th, 1842. On demand, for value received, I promise to pay James Hadley, or bearer, eight hundred^ and sixty-seven dollars and thirty-three cents, with interest after three months. Wm. R. Dodge. On this note were the following endorsements, viz. : April 16, 1843, received $ 136*44. April 16, 1845, received $319*. Jan. 1, 1846, received $518*68. What remained due July 11, 18^? Ans. $31*765 +. 3. $1000. Boston, Jan. 1, 1840. For value received, I promise to pay George A. Curtis, or order, on demand, one thousand dollars, with interest. Caleh Nelson. On this note were the following endorsements: — April 1, 1840, $24 ; AxLg. 1, 1840, $4 ; Dec. 1, 1840, $6 ; Feb. 1, 1841, $60 ; July 1, 1841 $4(ii^June 1 1844, $300; Sept. 1, 1844, $12; Jan. 1, 1845, $15 ; iSpct. 1 1845, $50. ;nmin( What remtined due, June 1, 1846 ? An$, $843*083 +. Note. 2 — Wh^n notes are paid within one year fironfthe time they SOS MSRCENTAGE. T 168. wem given, and have endorsements, it is common to subtract from the amount of the principal for the whole time the amount of ea^ ei|- dorsement ftasp. its date till the day of settlement. 4. $300. . MoWle, (Alabama,) June 10, 1846. For value received, we jointly and severally promise to Reuben Washburn to pay him, or order, on demand, three hundred dollars, with interest. Louis P. Legg. Sanford Comstock. On this note were endorsements : Jan. S!(), 1847, $116 ; Mardi 3, 1$47, $49*50 ; April 26, 1847, $85. What remained due June 2, 1847 ? Note. — The rate in Alabama is 8 per cent. Arts, $67*894 +. Connecticut Method. V 158, The Supreme Court of Connecticut have estab- lished a method somewhat different from the U. S. Court rule, which may be practised by those belonging to the state. The suBstance of the method is presented in the following I. Payments a year, or more than a year after the date of the note, or from each other, and those less than the interest due, are treated according to the U. S. Court rule. II. Find the amount of any other payment from date till one year from the time the note was given, or of a former cast, and subtract it from the amount of the principal for one year. The remainder is a new principal. Note. — If the note be settled in less than a year from the time of a cast, find the amount of^iich subsequent payment that has been made till tlie time of settlement, and subtract it from the amount of the principal found till tlie same time. $1100. Woodstock, Ct., Jan. 1, 1841. For value received, I promise to pay Henry Bowen, -^ order, eleven hu idred dollars, on demamd, with interest. "* James MarshalL On this note are the following endorsements: — Sept. I, 184 i, $30: April 1, 1842, $200; Dec. 1, 1842, $ 180 j^arch 1, 1844, $ 195 ; Sept 16, 1844, $ 250 ; May 16, 1846, $ icfl^uly 16, 1846, $170. ^^ What remains due Jan. 16, 1847? Ans. $234^34+. 1 159, 160 PERCENTAGE. 209 V 159. F(rr calmdating interest on a note in Vemumty payable at a specified time, with interest annually, on which payments have been nuide before it is due . From the amount of the debt found till the note is due, sub- tract the amount of the payments with the interest of each from its date till the same time. $800 • Townshend, Vt., Sept. 1, 1840. , For value received, I promise W. R. Ranney to pay him, or order, eight hundred dollars in ^ve years, with interest annually. Bushrod Washington. Endorsements: July 1, ld42, $200'; J&n. 1, 1844, $200; Sept 1, 1844, $300. What remained due Sept. 1, 1845 ? Am, (264 CoMPOTTND Interest. IT ICO. A promises to pay B ^5Q in 3 years, with in- terest annually ; but at the end of 1 year, not finding it con- venient to pay the interest, he consents to pay interest on the interest from that time, the same as on the principal. Simple interest is that which is allowed for the principal only. Connpound interest is that which is allowed for both prin- cipal and interest, when the latter is not paid at the time it becomes due. To calculate Compound Interest, Add together the interest and principal at the end of each year, and make the amount the principal for the next suc- ceeding year. From the last amqunt subtract the principal. 1. What is the compound interest of $256, for 3 years, at 6 per cent. ? — — ■ 11 I II ■■ l u ll » I I I I ■!■»■■■ M I Ml I ^ M I ■ ■■ — I ■■■■—■■■■ I .III II ■» ■ I ■ ■■ II ■ ■■■[■■■MILIUM ■ H I— ■■!■ ■ ^ Questions. — IT 160. What is simple interest ? — compound inter- est 7 What is th< method >f computing compound interest ? 18* 210 PERCENTAGE. IT 161 $256 given sum, or first principal. *06 271'36 amount, or principal for 2d year. *06 16*2816 compound interest, 2d year, ) added to- 271*36 principal, do. ) gether. 287*6416 amount, or principal for 3d year. *06 17*25846 compound interest, 3d year, ) added to- 287*641 principal, do. ) gether. 304'899 amount. 256 first principal subtracted. Atis. $48*899, compound interest for 3 years. 2. At 6 per cent., what will be the compound interest, and what the amount, of $1 for 2 years ? what the amount foT 3 years ? for 4 years ? for 5 years ? for 6 years ? for 7 years ? -— — for ,8 years ? Am. to the last, $1*593 +. IT 161* It is plain that the amount of $2, for any given time, will be 2 times as much as the amount of $1 ; the amount of $3 will be 3 times as much, &c. Hence, we may form the amounts of $1, for several years, into a table of rrmltiplicands for finding the amount of any sum for the same time. Questions. — ^ 161. How do we compute by the table? When I here are months and days, how do yon proceed? In what tune vnW any £ :m, at 6 per cent., double itself? inei. I^RCENTAGB. 211 Showing the amount of SI, or £1, for any number of years from 1 to 40, at 6 and 6 per cent., and from 1 to 20 at 7 and 8 per cent Tears. 1 2 2 4 6 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 Tears. 1 2 3 4 5 6 7 8 9 10 6 per cent. x-05 1*1025 ri5762-- 1*21550-- 1*27628-- 1*34009-- 1*40710-- 1*47745-- 1*55132-- 1*62889-- 1*71033-- 1*79585-- 1*88564-- 1*97993-- 2*07892-- 2*18287-- 2*29201-- 2*40661-- 2*52695 2*65329-1- 7 per cent. 1*07 1*1449 1*225043 1*310796 + 1*402551 -|- 1*50073 -1- 1*60578-- 1*71818-- 1*83845-- 1*96715-- 6 per cent 1*06 1*1236 1*19101-- 1*26247-- 1*33822-- 1*41851-- 1*50363-- 1*59384— 1*68947-- 1*79084-- 1*89829-- 2*01219-- 2*13292-- 2*26090— 2*39655— 2*54035-- 2*69277-. 2*85433 — 3*02559-- 3*20713-- 8 per cent. 1*08 1*1664 1*259712 1*360488- 1*469328- 1*586874- 1*713824- 1*850930- 1*999004- 2*158924- Tears. 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 6 per cent. 2*785963 2*925261 3*071524 3*225100 3*386355 3*555673 3*733456 3*920129 4*116136 4*321942 4*538039 4*764941 5*003189 5*253348 5*516015 5*791810 6*081407 6*385477 6*704751 7*039989 6 per cent. 3*399564 3*603537 3*819750 4*048935 4*291871 4*549383 4*822346 5*111687 5*418388 5*743491 6*088101 6*453387 6*840590 7*251025 7*686087 8*147252 8*636087 9*154252 9*703507 10*285718 Tears. 11 12 13 14 15 16 17 18 19 20 7 per cent. 2*10485-- 2*25219-- 2*40984-- 2*57853-- 2*75903-- 2*95216-- 3*15881-- 3*37993-- 3*61652-- 3*86968-- 8 per cent. 2*331638- 2*518170- 2*719623- 2*917193- 3*150569- 3*402614- 3*674823- 3*968809 - 4*286314- 4*629219- NoTB 1. — When there are months and days, first find the amount tor the years, and on that amount cast the interest for the months and days ; this, sidded to the amount, will give the answer. 1. What is the amount of cent, compound interest ? — '*50, for 20 years, at 5 per at 6 per cent. ? 213 PBRCENTAGE. T 162« Solution. — $ 1 at 5 per cent., hj like table, is $ 2*65339 ; there- fore, 2*65329 X 600*60 = $ J593*30 + Ans. at 5 per cent. ; and 3*20713 X 600*50 =1^ 1«26*881 + Ans. at 6 per cent. 2. What is the amount of $40*20, at 6 per ceat compound interest, for 4 years ? for 10 years ? for 18 years ? for 12 years ? ■ for 3 y;ears and 4 months ? for 20 years 6 months and 18 days ? Aru. to the last, $133*181+. Note 2. — Any sum at 6 per cent, compound interest, will dovble itself in 11 years 10 months and 22 days. 3. What is the amount of $750, at 7 per cent., compomid interest, for 16 years ? Am. ^2214*12. 4. What is tne amount of $150, at 8 per cent., compound interest, for 20 years ? IT 1^. Annual Interest. 1. $500*00. Keene, N. H., Feb. 2d, 1S&. For value received, I promise to pay George Hooper, or order, five hundred dollars, with interest anniiially till pai Henry Trui What was due Aug. 2, 1847, no payment having Psen made ? -^ Solution. —* A note like the above, with the promise to pt^ inter- est annually, is not considered, in courts of law, a contract for any- thing more than simple interest on the principal. Interest does not become principal by operation of law. If the annual interest be not paid, the creditor can bring his action for it at the end of each year. If he neglects to do this in Massachusetts, and in some other states, he is considered as waiving his claim, and must be contented to re- ceive simple interest. In New Hampshire, interest is allowed on the annual interest, in the nature of ddmages for its detention and use, ^m the time it be- comes payable till paid. Hence, when a note is written ** ^th inter- est annually," the following is the N. H. COUHT BUUEU Compute separately ^e interest on the principal, frdm the time the note is given till the time of payment, and the inter- est on each yew's interest from the time it should be paid, till the time of payment. The sum of the interests thus obtained % 1^. PERCENTAdtt. 213 will be the interest sought, to Tdiich add the principal ht the amount due. Applying this rule to a cast on the above note, the first year's interest of $30, is on interest 3 years and 6 months ; the ^ year's interest is on interest 2 years and 6 months, &c. The operation may be written down as follows : Interest on the principal, $600'00, 4 years 6 months, $135'00 1st year's int. ($30) 3 years 6 months, 6'30 2d " " 2 years 6 months, 4*60 3d " "1 year 6 months, 2*70 " 4th " . " 6 months, *90 (C u Amount of interest, $149*40 Then $600 + $149*40 = Am. $649*40, amount due. Note 1. — Among business men the mutual understanding and practice, oftentimes, is compound interest, when the note is written with interest annually ; but compound interest cannot be legally en- forced, unless it be so expressed in the note. The interest, by the method presented in the rule, is due at the end of each year, but as it is not paid then, it is on simple interest, just as any other debt would be, if not paid when due. Note ^.-^The same method is practised in Vermont when no pay- ments have been made, and there is no time specified when the note is due. (See 1 159.) 2. $1000*00. Brattleboro', Vt., June 10, 1842. For value received, we, jointly and severaHy, promise to pay Joseph Steen, or order, one thousand dollars, on demand, with interest annually. Samuel W. Ford, Stephen Wise. What was due Sept. 10, 1847 ? Ins, $1356*50. Note 3. -^ But when payments have been made, we find the amount of the principal for one year, and having found the amount of Questions. — % 163* When a note is written with the promise to pay mterest annually, how is it considered in courts of law? If the annual interest be not paid, what may the creditor do? If he neglects^ to do this, how is it considered by the courts in Massachusetts, and in* some other states? Wliat is the New Hampshire court practice? On what principle is simple interest allowed on the annual interest ? Re- peat the court rule. What mutual understanding among business men? How can the interest on interest be regarded as interest on any debt? What other state practises the same method? When payments ,have been made, what is done ? 214 P&RCENTAQE. t 163 each payment made daring that year from its date till the end of the same year, we subtract it from that amount, and the remainder will be a new principal, with which we proceed as before. 3. $400. Windsor, Aug. 2, 1844 For value received, I promise to Bishop and Tracy, to pay to them, or order, four hundred dollars, on demand, with an- nual interest Asa H. Truman. On this note are endorsements as follows : April 2, 1845, $ 50 ; June 2, 1845, $30 ; Jan. 3, 1846, $ 100 ; May 17, 1847, $80. What remained due Aug. 2, 1847 ? OPERATION. Principal, $400*00 Interest 1 year, 24*00 1st payment, $504- int. 4 mo. $1 == $61*00, ■■ . . 2d « $304-int.2mo. $*30 =30*30. Am V»^^ New prin. 342*70 &c. Ans, $194*34. The time, rate per ceTtt.t ^^ amoiwnX being giveny to find the principal. IT 163* 1. What sum of money, put at interest 1 year and 4 months, at 6 per cent., will amount to $61*02? Solution. — $ 1*08 is the amount of $ 1 for 1 year and 4 months, and $61*02 is the amount of as many dollars as the nuitber of times $1*08 is contained in $61*02. $ 61*02 -^$ 1*08 =s$ 56*50, the principal required. Hence, Divide the given amount by the amount of 1 dollar, at the given rate and time. 2. What principal, at 8 per cent., in 1 year 6 months, will amount to $85*12 ? Ans. $76. 3. What principal, at 6 per cent., in 11 months 9 days, will amount to $99*311 ? Qjiesiions* — ^ 163* What is the subject of this paragraph f Repeat the first example. Why do you divide $61*02 by 81*08? whdX is the rule for finding the principal, when the time, late per cent., and amount are given ? * f 164 PBRCENTAGE. 215 Note. — The interest of $ 1, for the ^yenr time, is '0564 « ^^U when there are odd days, instead of writing the parts of a mill as a common fraction, it will be more convenient to write them as a ilect' malj thus, '0565 ; that is, extend the decimal to four places. Ans. $94. 4. A produce buyer purchased 1000 bushels of wheat, on credit, and agreed to pay 15 per cent, on the purchase money; at the ejcpiration of 4 months he paid the debt and interest, which together amounted to $1500 ; what was the value of the wheat ? Ans. 81428^57 1 +. Discount. V 104* 1. I purchase a horse, agreeing to pay for him 8M06, one year from the time he comes into my possession, without interest. When I take the horse, I propose to pay down, for a just allowance ; what must I pay, money being worth 6 per cent. ? Solution. — I should not pay the whole sum, $106 j for the one who received it might loan it to a third person, and receive $112<36 for it at the end of the year, when he should receive only $106. Consequently, I must pay a sum, which, if loaned, will amount to $106 in a year, which is $100, Ans. An allowance made by a creditor to a debtor, for paying money due at some future time without interest, before the time agreed on for payment, is called Discount, and the sum paid is called the Present Worth, 2. I sell a piece of wild land in Wisconsin, for $868, to be paid in 4 years, without interest, since the purchaser is to re- *^ ceive no profit from his purchase. Wishing ready money, 1 transfer the debt to a third person for a sum, which, put at 6 per cent, interest, for the time, would amount to $868 ; what do I deceive for my debt ? Solution. — Since $1 in 4 years will amount to $ V24, for every Mfi $1*24 can be' subtracted from, or is contained in, $868, 1 shall recflfe «1. $868-j-$l'24 = $700, Ans, , The rule is e\4dently the same as in the last paragraph. Questions. — IT 164* What is the subject of this paragraph? Repeat the first example. SolTe it. What is discount? — presett worth ? How ia itf«niid? Pro^? now is dis«ouBt found? £16 PERCSNTAQB. 1[ 166 Pkooi^. — Find the amoimt (^ the resalt at the given rate and tkne : this amount will be equal to the given sum. 3. How much ready money must be paid for a note of $ 18, due 15 months hence, discounting at the rate of 6 per cent. 1 Ans. $16'744+. 4. What is the present worth of $ 56*20, payable in 1 year 8 months, discounting at 6 per cent, t at 4i per cent. ? af 5 per cent. ! at 7 per cent. 1 at 7^ per cent. T at 9 per cent. 1 Ans. to the last, $ 48*869. 5. What is the present worth of $ 834, payable in 1 year 7 months and 6 days, diwounting at the rate of 7 per cent. 1 Ans, $ 750 4-. 6. What is the discount on $ 321*63, due 4 years hence, discount- iog at the rate of 6 per cent.! Note. — If the present worth be subtracted from the given amount, the re- mainder will be the discount. Ans. $62*25+. 7. Sold goods for $ 650, payable one half in 4 months, and the other half in 8 months ; what must be discounted for present pay- ment, at 6 per cent. ? Ans. $ 18*8?2-)-. 8. A merchant purchases in New York city, goods to the amount of $ 5378, on 6 months credit, paying 6 per cent, mwe than if he had paid down. yWhat would he have saved if he had horr^vred ^money ^ ^ At 7 per cent, per annum, with which to make his purchase ! What would he save in 20 years, averaging 2^ such purchases each year? Ans. to the last, $ 6342. Commissiaii. IT 16S. 1. A merchant in Utica receives $988 from a house in New York, with which to purchase butter, after de- ducting for his services 4 per cent, on what money he shall lay out ; how much will he pay for butter ? Solution. — Of every $1<04 which he receives he must lay out 1 dol- lar, and retain 4 cents, thus having 4 cents for each dollar which he pays for butter. Then, as many times as $1^04 is contained in $988, so many dollars he must pay out. $988 -f- $1*04 s= $950, Ans. Note. — Had we multiplied $988 by *04 to ascertain how much he received #9r his services, it would nave given him 4 cents on each dollar which he re- ceived. This would have given him 4 cents for laying out 96 cents, inst«»M of 100 cents, as required by the question. Questions* — IT 165* What is the first example ? Give the solution. Show the error, if performed as supposed in the note. What is commission 1 What are persons, who buy and sell goods for others, called 7 When they receive money to disburse, now is their conmiissia^ estimated 7 When m value of die gdodn bought or sold is known, how is the oonmiission etti- matwl? * > f f 166. PERCENTAGE. 217 The per cent, or amount allowed to persons for their ser- vices in assisting merchants and others in purchasing and selling goods, and for transacting other business, is called Commission, Persons who buy and sell goods for others, are called Agents^ Commission Merchants, Corrtspondents, and Factors, When agents, &c., receive money to disburse, their com- mission is estimated in the same manner as discount, by the rule H 163. 3. Received $ 2475, with which to piudiaae wool, afler deduct- ing my commission of 5 per cent. ; how many dollars will I pay out! What will be the amount of my commission f Ans, to the last, $ 117'857. 3. Sent my&gent $4820, with which to purchase wheat, afler deducting his commission of 7^ per cent. ; how much money will he expend, and what will be the amount of his commission ? Ans. He will expend $4483'72-f-- NoTB. — When the value of the goods bought or sold is known, the com- mission is estimated upon that value, in the same manner as percentage. (IT 143.) 4. A commission merchant sold goods to the amount of $ 1422, at 5 per cent, commission ; how much did he receive for his services? Ans. $71*10. 5. My correspondent sends mb word that he has purchased goods to the value of $ 1286, on my account ; what will be his commission, t 21^ ner cent. ? Ans. $ 32' 15. 6. What must I allow my correspondent for selling goods to the amount of $ 2317*46, at a commission of 3^ per cent. ? Ans. $75*317. 7. A tax on a certain town is $ 1627*18, on which the collector is to receive 2J per cent, for collecting ; what will he receive for col- lecting the whole taxi Ans. $40*6'/9. The time, rate per cent., and interest being given, to find the principdL IT 166. 1. What sum of money, put at interest 16 months, will gain 810*50, at 6 per cent. ? Solution. — $1 in 16 months, at 6 per cent., will gain $'08; and since $<0d is the interest of $1 at the given rate and time, $10<50 is the interest of as many dollars as the number of times $<08 is contained in (can be subtracted from) $10<50. $10'50-i- $'08 =» $131*25, the prin- cipal required. Hence, the Find the interest of 81, at the given rate and timet by 19 218 PERCENTAGE. % 167, 16& whicH divide the givsn interest; the quotient will be the prin- cipal required. BXAMPLJBS FOR PRACTICB. 3. A man paid $4*5Q interest, at the rate of 6 per cent., at the end of 1 year 4 months ; ^hat was the principal t Ans. $56*50. 3. A man received, for interest on a certain note, at the end of 1 year, $ 20 ; what was the principal, allowing the rate to hare been 6 percent.? Ans, $333'333^ 4. A man leases a farm for $562, which sum is 10 per cent, of the value of the form ; how mnch is the farm worth? The princvpcH^ interest , arid time being given, to find the rate per cent. IT 167. 1. If I pay $3*78 interest, for the use of $36 for 1 year and 6 months, what is that per cent. ? Solution. — The interest of $36. at 1 per cent,, for 1 year and 6 months, is $'54, and consequently $3 '78 is as many per cent, as the times $'54 is contained in $3'78. $3'78 -i- «54 = 7 per cent. Hence, the RULB. Find the interest on the given sum, at 1 per cent., for the given time, by which divide the given interest; the quotient will be the rate at which interest is paid. KXAMPLil» FaR PRACTICE. 2. If I pay $2*34 for the use of $468, 1 month, what is the rate per cent. ? Ans, 6 per cent. 3. At $ 46*80 for the use of $ 520, 2 years, what is that per cent. ? ^ Ans, 4i per cent. 4. A stockholder, who owned 10 shares, of $ 100 each, of the Tonawanda Railroad company stock, received a dividend of $ 50 every 6 months ^ what per cent, was that dn the money invested ? Ans, 10 per cent. 5. A Widow lady, whose expenses are $ 324 a year, has $ 5400 in money : at what rate per cent, must she loan it, that the interest may pay her expenses? The princvpalj rate per cent.y and interest being given, to find the time, IT 168. 1. The interest on a note of $36, at 7 per cent, was $3*78 ; what was the time ? Questions* — IT 166* When the time, rate per cent., and interest are given, how may the principal be found 7 Explain the principles of the rule. IT 16T. When the principal, interest, and time are given, how do you find the rate per eent. ? Explain the prmciples of the rule. Tiflb. PSROBNTAOB. ' 219 SdumoM. — Hie interest of $36. 1 year, at 7 per cent, is $2<52. Since ♦2*52 will pay for the use of $36 1 yau-, $3'78 will pay for the use cf it as many years as the times $2<52 is coutaihed in $3^. $3^8 -i 2<52 -^l^^ years, as 1 year 6 months, the time required. Hence, tke « Find the interest for 1 year on the principal given, at the given rate, br which divide the given interest ; the quotient will be the time required, in years and decimal parts of a year. sxAMPiiEs f6r practice* 2. If $ 3l'7l ihterest he piaid on a note of $ 236-50, what was the time, the r^te being 6 per ceikt. ? Ans. 2^^ es 2 years 4 months. 3. On a note of $600, paid bterest $20, at 8 per cent. ; what was the timet Ans. *416-(-yr. = 5mo.,nearly. It would be exactly 6 but for die fraction lost. 4. The interest on a note of $ 217'25, at 4 per cent. , was $28'242 ; what was the timel Ans, 3 years 3 months. NoTB. — When the rate is 6 per cent., we may divide the interest by i the principal, and the quotienL removinc the separatrix two places to the lefl^ will be me answer requirea, in months and decimals of a month. The percentage of any number of dollars being given^ ^ Jind the rate. ^ IT 169« 1. A merchant purchases a piece of broadcloth fos $60 ; what will be the per cent, of gain, if he sells it for $67*20 ? $67'20 eO'OO Solution. — Subtracting the price which he gave 60* ) 7*20 ( *12 fTota the price for which he sells the cloth, we have 60 ^®^ $7*20, the gain, on $60, of which we must take ^ ^ for the gain on $1. "We get 12 cents as the gain on $1, or 100 cents. Hence the gain is 12 hundredths 1*^0 of the sum paid, or, Ant. 12 per cent. 120 Hence, the Divide the percentage of the number of dollars by the num- III ' • ' '■ ■ Questions*— IT 168* When the principal, rate per cent., and interest e given, how do you find the time 7 £zplau the principles of th« rule. 220 PERCENTAGE. T 170 ber of dollars on which it has accrued ; the quotient, which is the percentage of SI, or 100 cents, will express the rate per cent. SXAMPI^ES FOR PRACTICE. 9. A merchant purchases goods to the amoant of $ 550 ; what per cent, profit must he make to gain $ 66 ? Ans. 12 per cent. 3. What per cent, profit must he make on the same purchase, to gain $38'50? to gain $24*75? to gain $2<75? Ans. to the last, '005, or J per cent. 4. Bought a hoffshead of rum, containing 114 gallons, at 96 cents per gallon, and sold it again at $ 1*0032 per gallon ; what was the whole gain, and what was the gain per cent. ! A s \ $^'^^1 ^^^ whole gain. ' ( 4i, gain per cent. 5. Bought 30 hogsheads of molasses, for $ 600 ; paid in duties $20'66 ; for freight, $40*78; for porterage, $6*05, and for insur- ance, $ 30*84 ; if I sell them at $ 26 per hogshead, how much shall I gain per cent. ? Ans. 11*695 4- per cent. 6. A ^>endthrift, who received an inheritance of $3000, spent $ 960 the first week in gambling ; what per cent, of his money is gone! Ans, 32 per cent. 7. A farmer paid $ 2*50 for insuring buildings worth $ 1000 ; what was the rate per cent. ? Ans, i per cent. 8. A commission merchant receives $ 37^50 for selling goods to the amount of $ 1250 ; what was the rate per cent. ! Ans. 3 per cent. ' 9. A broker receives $ 270 for selling $ 18000 worth of stocks ; what is the per t^nt. for brokerage ! Ans, H per cent. 6^ Bankruptcy. IT 170« An individual who fails in business sometimes makes an assignment of his property, which is divided among his creditors according to their respective debts. In making calculations in bankruptcy, we find what can be paid on each dollar owed, and multiply this by the number of dollars which each man claims, to get his share. EXAMPLES FOR PRACTICE. . 1. An extensive banking house in New York iails for $800,000 » the property of the house is found to be $ 300,000 ; what is paid oa a dollar 1 Ans, $'37||, or 37^ per cent. 2. How much will a man receive on the above, whose dues ate $16500! Ans. $6187'50. l^vesHoiis.- T169* How U Ex. 1 explained 7 Role. 1 171. PERCmTAGO. 281 3. A mereltant fafls in Ininness, owing to A, $350 ; to B, $ 320 ; •oC, $500, to D, $180; to {:, $700; to F, $390; to G, $65*50; jo H, $1300; to I, $2200; to J, $850; his property is found to be $ 4653 ; how much does each receive T General Average. V 171 • When a ship is in distress, the expense incurred, or the damage suffered by the ship, or any part of iSe cargo, i» averaged upon the value of the ship ; upon the cargo, esti- mated at what the goods will bring at the destined port ; and upon the freight, d^ucting one third, on account oi the sea- men's wages. • To estimate general average^ we find the proportion of the loss on each dollar, and then the loss on the number of dol- lars of each contrlbutary interest. The ship Silas Richards, in her voyage from New York to Charles- ton, became stranded on the coast of North Carolina, when it was foond necessary to throw overboard 506 barrels of flour, belonging to Goodrich & Co., worth $6*87 per barrel. The expense of getting the vessel off, was $ 197 ; of supplying new rigging, $240, of which one third is to be deducted, as the new is supposed to be better than the old. The ship is worth $10232; the freight is $4800, of which one third is to be deducted. Goodrich & Co.' had on board 1000 barrels of flour; M. H. Newman & Co., goods worth $4000 ; D. Appleton, goods* worth $ 5236 ; Hyde & Duren, goods worth $ 9000 ; how much do Groodrich & Co. realize for all their ^our ; what does each interest contribute towards the loss, and what is the rate per cent, on the contributary interests? Goodrich & Co. realize $6186*669. The ship's portion of the loss is $1017*735 -4-. Portion of the freight, $318*291 - -. " M. H. Newman & Co., $397*863--. ** D. Appleton, $520*803--. " Hyde & Duren, $895* 193 - -. " Goodrich & Co., $683*331-}-. Rate per cent., 9|f M* Questions* — IT 170« What do you understand by bankruptcy? How are calculaiions in bankruptcy made? IT 171. What do you understand by general average ? What expenses and losses may it embiace? On what rifferent interests are tae expenses averaged? .19* B88 Maosm'AOB. T ITS. Partn^Fshlp- V 173* When two or more persons unite a part or the whole, of their capital for the- prosecution of businesj, they form a Company or Firm, and their business is called Partner* ship business. Each member of a 6rm is called a partner. Capital or Stock is the money or other property employed in trada; Joirit Stock is stock of a company or firm. Divi' dend is the gain or loss to be shared among the partners. 1. Two persons have a joint stock in trade ; A put in 9250, and B $350 ; they gain $1^ ; what is each man's share ? OPEBATION. Solution. — We di- 6|00W0'00 ?i^1,,S''60S;';^ch^li ^^- give us the gain on $1, W26 as $^. Then 250 times A's gain, -25 X 250 = S62*50, > . «*25 is A's ^in, and 350 m gain, 25 X 350 = $87*50, ( '^^' times $<25 is B's gain. ° ^ ' By the first cmeratiott, we get the rate per cent, of gain or loss, according to ^ 169. The sec- ond may be performed by the ordinary rule for percentage. Or the operation may be performed as follows : A's gain tmllbe^ = ^o( $150 = $62*50. JB'* gain wiabeUi = ^of $150 = $87*50. Hence, Take such a part of the whole gain or loss as each man's stock is part of the whole stock ; the part thus taken will be his share of the gain or loss. NoTjB. — This rule may be applied to tne operations in several preceding paragraphs. SXAMPIiES FOR PRACTICE. 3. A, B, and C trade in company ; A's capital was $ 175, 6's $200, and C's $500 ; by misfortune they lose 250 ; what loss must each sustain ? ( $ 50, A's loss. Ans, I $ 5ri42f , B's loss. ( $142*857J, C's loss. 3. Divide $600 among 3 persons, so that their shares may be to each other as 1, 2, 3, respectively. Ans. $ 100, $ 200, and $ 300. i^nesttoiLS. — IT 172. What is a compimy or firm 7 . — partnership busi- ness ? — a partner ? — capital or stocL ? —joint stock 7 — dividend ? Giv« the first solution of Ex. 1 ; — the se^ .nd. Rule. T 173, PERCSNTAQE. 223 4. Two meTchaats, A aod B, loaded a ship with 500 hhds. of ram; A loaded 350 hhds., and B the rest; in a storm, the seamen were obliged to throw overboard 100 hhds. ; how much mu$t each sustain of the loss ? Ans, A 70, and B 30 hhds. 5. A and B companied ; A put in $ 45, and took out f of the gain ; how much did B put in ? Ans, $ 30. Note. — They took out in the same {Npoportion as they put in ; if 3 fifths of the stock is S 45, how much is 2 fifths of it? 6. A and B companied, and traded with a joint capital of $ 400 ; A received, foT his share of the gain, ^ as much as B ; what was the stock of each! <f ! / ■ ■ Z^ a , S ^ 133*333J, A's stock. V- jLns. ^ ^366*6661, B's stock. 7. A and B ventured equal stocks in trade, and cleared $ 164 ; by agreement. A, because he managed the concerns, was to have $ 5 of tl^ profits, as often as B had $ 2 ; what was each one's gain ? and how much did A receive for his trouble? / / / c / ' Ans, A's gain was $liri4df, and B's $46'857|/and A re- ceived $70*285n^ for his trouble. 8. A cotton fectory, valued at $ 13000, is divided mto 100 shares; if the profits amount to 15 per cent, yearly, what will be the profit accruing to 1 share? to 2 shares i to 5 shares ? to 25 shares? Ans. to the last, $450. ' 9. In the above-mentioned factory, repairs are to be made which will cost $ 340 ; what will be the tax on each share, necessary to raise the sum? — on 2 shares? — — ^ on 3 shares? « on 10 shares? Ans, to the la9t, $34. 10. Two men paid 10 dollars for the use of a pasture I month ; A kept in 24 cows, and B 16 cows ; how much ^ould each pay ? Paetnershif on Time. IT ITS* 1. Two men hired a pasture for $10 ; A put in 8 cows 3 months, and B put in 4 cows 4 months ; how much should each pay ? Solution. — The pasturage of 8 cows for 3 months is the same as of 24 eows for 1 month, and the pasturage of 4 cows for 4 months is the same as of 16 cows for 1 month. The shares of A and B. therefore, are 2^4 to 16, as in the former question. Hence, « When time is regarded in partnership, multiply each on^s stock by the time he continues it in trade j and use the product for his share. Ans. A 6 dollars, and B 4 dollars. 2. A and B enter into partnership ; A puts in $ 100 6 months, and then puts in $ 50 more ; B puts in $ 200 4 months, and then takes out $ 80 ; at the close of the year they find that they have gained $95; what is the profit of each? .^ j $43*711, A's share. ^'**' } $51*288, B's share. Questions* — IT 173. What are we to understand by partnership oi time ? How do we proceed ? 1^ 2M PBRCENTAGEL Y 174.' 3. A, with a capital of $500, began trade Jan. I, 1846, and, meeting with success, took in B as a partner, with a capital of $600, on the ifst of March following ; four months after, they admit C an a partner, who brought $ 800 stock ; at the close of the year, the« find the gain to be § 700 ; how must it be divided among the part ners^ ( $250, A's share. Ans. { $250, B's share. ( $200, C'sahare. Banking. ^ 174* A bank is an incorporated institution which traf* fics in money. Bank notes, or bank bills are promissory notes, payable to bearer. Banks loan their money on notes, the interest always being paid in advance. For example, B holds A's note for $100, payable in 90 days, without interest. But B is in immediate want of money. He carries A's note to a bank, and if their credit be un« doubted, the bank will receive A's note and pay the face of it, minus the interest on it, for 3 days more than the given time, (90 -[- 3 = 93 days.) The note is then said to be discounted at the bank. These 3 days are called days of grace. But the bank will require B to write his name on the back of the note. This is called endorsing the note. It subjects B to pay the note when the 90 days* and the 3 days of grace shall expire, provided A, who gave the note, should fail so to do. The money received from the bank for the note, is called the avails of the note. The note is said to be mature when the time that it should be paid shall arrive. Again : B as principal, with C and D as sureties, may give their note jointly and severally to the bank, for the sum tv^anted, payable at a specified time, without interest. Then if B fails to pay the n9te, his sureties, C and D, either or both, will be holden to pay it. $100*00 Principal. Note. — Bank interest, when the rate is 6 ^r $1*00 Int. 60 days. cent., may be <»st by in- tcn « OA J specUon, as follows : Let *50 JO days. tJe sum on which it is to *05 " 3 days of grace. be cast, be tlOO. The principal itself. remoTing the pomt two places to thi $1*55 Discount for 90 days and grace. left, is made to express the interest for 66 days, (91*00 ) half of which (fSO) is the interest for 30 day% T 17& PERCEI^AQB. 226 and -j^ of the interest for 80 days, rso-f-lOsros, is the interest ftr i daTs. iThe sam of these results is the bank interest on 9100, at 6 per cent., for 90 days, which sum 9i<55, deducted from the face of the note, makes its avails to B, •f8'45. If the discount be other than % per cent., take such fractional part of the dis- count at 6 per cent, as the required rate is less or more than 6 per cent., which added to or subtracted from the discount at 6 per cent., as the case may re- quire, will give the discount sought. BXAMPIiKS FOR PRACTICB. 1 . What win be reeeived on a bank note of $ 500, due in 90 daySy at 7 per cent.? , Ans. $490*55|. 2. What is the diacoont of a bank note for $ 300, due in 00 days, at 5 per cent. ? Ans. $ 3*875. 3. What is the discount of a note for $600, due in 90 days, at 8 percent.? Ans. $12*40. 4. What is received on a note for $ 740, due in 90 days, at 6 per cent.! . Ans. $728*63. 5. A man gets a note of $ 1000 discounted for 90 days, at 6 per cent, per annum, and lends the money immediately, tul the time when he is obliged to pay : what does he lose? Ans. $ *244'. Taxes. IT 17ff« A tax is a sum imposed on an individual for a public purpose. ^ A Poll tax is a specific sum assessed on male citizens above 21 years of age ; each person so assessed is called a poll. Taxes are usually assessed either on the person or prop- erty of the citizens, and sometimes on both. rroperty is of two kinds, personal property and real estate. Personal is movable property, such as money, notes, cattle, furniture, &c. Real estate is immovable property, such as lands, houses, stores, &:c. An Inventory is a list of articles. In assessing taxes, it is necessary to have an inventory of all the taxable property, both personal and real, of those on whom the tax is to be levied, and also, (if a poll tax is to be raised,) of the whole number of polls ; and as the polls are Questions. ~ IT 174. What is a bank? When is bank interest paid? Illustrate by the example. What is meant by days of ^race? How long is the interest cast on a note due in 90 days ? What security does the bank re- quire ? How is bank interest found bv inspection, when the rate is 6 per etnt. ? How, when it is any other rata 7 . < « 226 PfiBGKNTAOil. ^176. rated at a eertain sum each, we must first deduct from tbe whole tax the amount of the poll tax, and the remainder is to be assessed on the property. The tax on $1 is found by dividing the amount to be ^- sessed on the property by the value of the property taxed. The tax on ^ny anuncnt of property is found by multiplying the value of the property by tne tax on SI. 1. A tax of 9917 is to be assessed on a town in which are 320 polls, assessed at 40 cents eac}i ; the iuFentoried v^lue of the personal and real property of the town is $5260Q ; what is the amount of the poll taxes ? How much remains to be assessed on the property ? What is the tax on $1 ? Solution. $ <40X 320 aB$128, amount of poll tax; then $917—9128 ss$789. amount to be i^ses^ on property^ and $78^ -f- $52600 sss $'015, Uie tax on $1. NoTB. — In making out a tax list, form a table eofitainiag the taj^es 09 1 , 2, 8. &c., to 10 dollars: then on 20. 30, &c., to lOO dollars; and then on IQO, 200, »c., to 1000 dollars. Then, knowing the Inventorv Of any individaal| it is easy to 0nd the tax upon his property. Let us apply this method in assessing a tax on a town. 2. A certain town, valued at $64530, raises a tax of $2259*90 ; there are 540 polls, which are taxed $*60 each ; what is the tax on a dollar, and what will b^ A's ta^i whose real estate is valued at $1340, his personal property at $874, and who pays for 2 polls ? SoLXTTioN. 540 X ^ '60 == $324, amount of the poll tsxes, and $2259<90 — $324ssl935<90, to l)e assessed on property, and $1935<90 -^ $64530 s $<03, tax on $1. TABIifi. dolls, dolls. dolls, dolls. doUs. dolls. Tax on 1 is <03 Tax on i 10 is '30 Tax on 100 is 3< if 2 « *06 i< 20 " <60 « 200 « 6' it 3 « ^09 « 30 " '90 « 300 " 9< u 4 " a2 C( 40 " 1'20 50 " 1'50 « 400 " 12' it 5 « '15 (( « 500 " 15' u 6 « '18 u 60 « 1'80 « 600 " 18' u 7 « '21 u 70 " 2'10 •i 700 " 21' u 8 « '24 u 80 " 2'40 « 800 " 24« u 9 « '27 u 90 " 2'70 « 900 " 27' " 1000 « 30* Qnesti^MUk— 7 175. What is a tax 7 — a poll tax? How are taxes usually assessed 7 ' Pro])erty is of what kinds? What is personal property? — real estate? What is ^ inventory? In assessing taxes, what is to oe done ? When a poll tax is to be raised, what must first be done ? How do you find the amount to be assessed on the property ? How do you find the tax on tl ? How on any amount of property ? What course is usually pursued by assessors m makmg out a tax list 7 Explain the manner in which any in- dividual's tax is made out firom the assessors lax tab.e. How may a tax list be proved? T179. PERCENTAGE. 2Spt Now to find A's tax, hb teal estate teing $1340. we find, by tlie table, that The tax on . . $1000 • • is • . $30< The tax on , . 300 . . " . . 9' The tax on . . 40 . . " . . 1*20 Tax on his real estate, $40<20 In like manner, we &id the tax on his pemmal p ro pe rty to be • 26*22 2 polls at *60 each, are ... . . . . 1'20 Amtnmty $67<62 3. What will be the amount of B's tax, of the same town, whoee inventory is 874 dollars realy and 310 dollars personal property, and who pays for 3 poUs 1 Arts, $34'32. 4. of C's pajdng for 2 polls, whose properly is valued at $34821 --< — of D's, paying for 1 poll, whose pioperty is valued at $4675? Ans. to the last, $140'85. Proof. — After a tax list is made out, tuid together the taxes of all the individuals assessed ; if the amount is equal to the whole tax assessed the work is right. Duties. IT 176* Duties or customs are taxes on imported goods. A custom-house is a house or an office where customs are paid. Government has established a custom-house at every port in the United States into which foreign goods are imported. Besides duties on merchandise, every vessel employed in commerce is required to pay a certain sum for entering the ports. This sum is in proportion to the size or tonnage of the vessel. The income to the government, from duties and tonnage, it* called Revenue. All d^ties are imposed and regulated by the general gov- ernment, and must be the same in all parts of the Union. Note. — A table of the duties imposed by government is called a Tarifl*. The law requires that the cargoes of all vessels engaged ^n I I II 1 1 I III. Qaestipns. — IT 170* What are duties ? — costom-houses, where, and bj whom established ? What tax is named besi^les duties, and how proj^r- tumed ? What is revenue ? How are duties imposed ? What is a tariffl How b the value of the gfoods in a vessel ascertained ? How many kinds of duties? What are specific duties? — ad valorem duties ? Defiue ad vilo- fem. MHHHHHMHlBaHAM PERCENTAGE. IT 177. foreign commerce, shall be weighed, measured, or gauged, by the custom-house officers, for the purpose of ascertaining the amount or value of the goods on which duties are to be paid. Duties on imported goods are of two kinds, Specific and Ad Valorem. . A Specific duty is a certain sum per ton, hundred weighty pound, hogshead, gallon, square yard, foot, &c., without re- gard to its value. Ad Valorem signifies upon the value. An Ad Valorem duty is a certain per cent, on the sum paid for the goods in the country from which they are brought. Specific Duties. IT VT7* In the custom-house weight and gauge of goods, certain deductions are made for the box, bag, cask, &c., con- taining the goods, and also for leakage, breakage, 6cc. These deductions must be made before the specific duties are im posed. Gross weight is the weight of the goods together with the box, bale, bag, cask, &c., which contains them. Draft is an allowance made for waste, which is to be de- ducted from the gross weight, and is as follows : On 112 lbs. 1 lb. Above 1 12 lbs., and not exceeding 224 lbs., 2 lbs. " 224 lbs., " " 336 lbs., 3 lbs. " 336 lbs., " " 1120 lbs., 4 lbs. « 1120 lbs., " " 2016 lbs., 7"lbs. " 2016 lbs. 9 lbs. Tare is an allowance for the weight of the box, bag, cask, ice. It is to be deducted from the remainder of any weight or measure, after the draft or tret has been allowed. Leakage is an allowance of 2 per cent, on all liquors in casks, paying duty by the gallon. Breakage is an allowance of 10 per cent, on ale, beer, and portei in bottles, and of 5 per cent, on all other liquors in bot- tles ; or the importer may have the bottles counted, and pay duties on the number remaining unbroken. Questions* — IT 177* What deductions are made, if goods pay specific duties 1 What is gross weight ? — draft, or tret ? How much is it, and when deducted? What is tare, and when deducted? What is leakaoe? — - breakage, and wnat privilt^e has the importer ? — net weight ? R«d« far oil* eolating specific duties. '* T !?&• PEBCBNTAQB. SS9 Net weight is the weight of the goods after deducting the weight of the box, bale, tec,, and making all other allow- ances. 1. M'lat is the specific duty on 5 hogsheads of molasses, each containing 120 gallons, at 12^ cents per gallon, the cus- toniary allowance being made for leakage ? SoT.oTTOi' Since there are ISO gallons in 1 hogshead, in S ht^s- heads Acre are 5 times 120 gallons = 600 gallons. 2 per cent, of 6U0 gallons is eOO X '02 = 12 gallotis. and GOO gallons— 12 gallonsn 5BS gallonati^ince ihe daly on 1 gallon is 13j ceais, the dnty on S8e g*^ Ions is ^ limes t'125 = t73'50. Hence, To find the specific duty on any given quantity of goods, * RULE. r* Deduct from the given quani lowance for draft, tare, leakage a II. Multiply the remainder by weight or measure of the goods, a duty required. EXAMPLES FOR S. What is the speciiic dnlj on 75 83 pounds gross, tare in the whole 3. What is the duty on 430 dozen hi bottle, the costomBTj allowance being i 4. What is the dnty on B hogshead ewi. Q qra. gross, tare 14 Ibe. per cwt.. 5. What is the duty on 1 barrels weighing 171 pounds rtobs, the seoon 109 pounds gross, and the fourth 99 pounds gross, at S^ cents pet pound, the customary allowance being mode for draft, and 13 pounds per barrel for tare e Ata. f 27-25. Ad Valorem Doxies. H 178. Since od valorem duties are estimated upon the ictual coqt of the goods, it is plain that they are found by simply multiplying the cost of the goods 1^ the given per 930 PEBCENTAGE. ^T 179- Nora. — In MllnuliDS ad laloreiii dmies, no deducloiu of iiit nind ue to bamadc. 1. What ia the ad valorem duty, at 25 pt r cent , on 32 yds. of English broadcloth, which cost «4'75 pe: yard? SoLDTCoti. — 32 yds, a! H'7S, cost 4'75 X 32=1152, and 25 pel oem. of 152, is 138, Am. HXAMPLES FOR FHACTICE. 3. What ia the ad Talorem dutj', at 18 per cent., on 4^ bags <^ Javo coSee; each weighing 115 pounds, and which ooM Hi cents per pouiufT ^Ji»jp3'16. 3. What is the ad Taloiem duty, at 33^ pra cent., otagiom of Sheffield cutlery, which cost *S56'80! JnlWeS'tiO. 4. Whalia the duty, at SO percent., on a piece of Tutt^ carpet- ing, coi.rwning 140 yards, and which cost $l'»3 per yajpt? What ia the duly on 1 yatd! For how Auch must it be sold per yard, to die cost and duty ! Ans. to the last, $i*«B. duty, at 35 per cent., on a case of Itijl&n silks, U Aru. $\ti8T&2l,. duty, at 15 per cent., on 3 dozen gold watches, ,1 Jnj. $367'20. iity, at S3 per cent., on 75 cAeala of tea, the net : being 03 pounds, and the tea costing 41 cents Aiu. $623'38 I. Review of Percentage. hat is meant by percentage ? — rate per cent. ? — are calculated by percenloge? What is insurance? I What is meant by stocks f — brokerage r — ,..w«. - — .. .. aat is interest, and how calculated! How is 6 per cent, interest calculated by inspection ! What is meant by partial pay- ments? What diSereuce between the U.S. and Conn, rates T To what case iloei the Vt..rule for partial payments apply f What is done when notes with partial payments are paid within a year! How does com- pound di&er from annual interest ! Like what ca.se in interest are dis count and commission? What do yoti say of baakraplcy? —of general average? — partnership! How does the calculation of bank interest differ from that of other interest! How are lales assessed? How do dutif-s dilfer from other taxes? What two ki'ils of dnties, and how is each computed? 1. ^Whal is the interest of $373'51, for year and 10 days, ai 7 percent.! Ans. $lQ-*iT7 'i-'. 9. What is the inteieet of $466, for 1 year 3 months 10 days, aj 8 pet0OTt.1 Xn». fso-ess. 1 17»^ PERCENTAGE. 231 S. What ia the inUrcM of f I6D0, for 1 yew and 3 monthe, at 6 per cent. ? Am. $130. 4. Whii IB the intaest of $5*811, fw 1 jeai 11 moalhs, at 6 pei eeut.l Am. $'66B. 5. What is the inlereat of ^99, fbi 1 mipth 19 dajB, at 3 per oent. 1 Am. '000. Am. |s-560. 7. What is the intereBt of 9IT>68, foi H iniiDthH28 days, at 6 pei cent. T Am. $1'054. 8. Whatistbeinleiratof $200,for 1 day, atO percent.! 9 dajm? 3 daya! i daya? 5 daya! Am. for 5 days, SO'ISO. O. W>at ia theintereBt ofhalfamil], for567 yean, al 6 per cent. ! Am. won. 10. What is the intereat of $61, for 3 years 14 days, at i per cent. ! i per cent. ' | per cent. 1 8 per cent. ! 3 pel cent ! 41 per cent. ! 5 per ce 7 per cent. 1 7i per cent. ? 8 eent. ? 10 per cent.1 13 per cent.? 11. What is the interest of S cents, for 45 y «t 6 per cent. ? 12. A'a note of $175 wae given Dec. 8, i: doTsed one year's intereat ; what waa there du at 7 per cent. ? 13. B'b note of $S6'75wa8 given Jane 6, per cent., after 90 days ; what was there due I ' 14. C'b 15. D's note of $303'17 was given Oct. 5, Kr cent. atW 3 months ; Jan. 5, 1839, he pai< eMay Q, 18411 16. E'b note of $B70'05 was given Nov. 17, 1840, on intereat at 6 per cent, after 90 days ; Feb. 11, 1845, he paid $186'06 ; what was there due Dec. 33, 18471 Am. tl041'5e. 17. Supposing a nolB of $3I7'B3, dated July 5, 1637, on which were endorsed the following payments, viz., Sept. 13, 1839, $S08'04 ; March 10, 1840, $76 ; jwhai was there due Jan. 1, 1841, interest at 7 percent.? Am. $93'032. le. What will be the annual msurance, at | per cent- on a house valued at $1600^ .t ns. «10. 19. What will be the insurance of a ship and carg; , valued at $5643, at IJ per cent.! at J per cent.1 « ^^ jfM •eat. ! at y- per cent. 1 at | per cent. ? Am. at t per cent. $4S'399. 833 PERCENTAGB. ^T 179. 50. A man having compromised with his creditors at 62} cents on a dolki, what must he pay on a debt of $13T'46) Ans. tS5'9l3. 51. What istlie va1ueof$S00 stock Id the Utica and Schenectady railroad, at lISj per cent. T Ans. *1>00. 22. What is the TsJue of tS60<T5 of stock, at 93 |>er cent. ? Ans. *521'407. 23. What principal, at 7 per cent., will, in months 19 days, amount to $422'401 Ans. $400. 24. What is tlie present worth of $426, payable in 4 years and 12 days, discounting at the rate of 5 per cent. ^ In large sums, to bring out the cents correctly, it will sometimes be necessary lo extend the decimal in the divisor Ut fire places. Arts. $3r>4'507-f. 25. A loecchant purchased goods for $250 ready money, and sold them airain for $300, payable in months ; what did he pan, dis- 1.? Ans. $3T'081. • ir S3120, to lie paid, one half in 3 months, and intlis ; what must be discounted for present pay* Ans. $6e'491 -f . an a certain note, for 1 year S months, at 6 pftr what was the principal? / ' Ans. 9i75. aJ, at 5 per cent., in 16 months 24 days, will \ ( Am. $500. iO interest for the use of $500, 9 months and 9 < per cent, ? esat £'lCTper lb,,and.;KU them at 20 cents, aying out SlOOl ' Ans. $19-76. Ions ol' brandy, at $l'ro per (tallou, and sold it ained or lost par cent. ? at $4-49 pi^ yard ; how must I sell it to gain Ans. «5'04. Ims ul' brandy, at 93 cents per gallon, but by laked out ; for what must I sell the remainder on iho whole cost at the rate of 10 per cent. ? Alts. $1-265 per gallon. 34. A merchant bought 10 tons of iron for $950 ; the freight and duties were $145, and hia own charges $25 ; how must he sell it pet lb. to ^in 20 per cent. 1 Ans. 6 cents per lb. 33. A note is given for $2000, at per cent, annual interest, pay- able in 6 years; the date of the note fs Dec. 1, 1841 ; there are en- dorsements upon it aa follows : June 1,^913, $163 ; Feb. 1, 1843, $12 : Jan. I, 1844, $300; April 1, 1845^$20; June 1, 1845, SSO; Aug. 1, 1845, $400; Jan. I, 1846, $100; Aug. 1,1847, $150; Oct. 1, 1847, $75. What remaned duc,'T)ec. 1, 1847, calculated by the U. S. Court rule, l-y the Conno^cut rule, and by the Ver- moat rule, and how do the results compare? Ant U. S., S.1368'81 ; Ci:, $1373'56 ; Vu, 1274'7e. <^ 1 180, 181. KaUi^TlON OK PAYMENT?. 233 EQUATION OF PAYMENTS. iri80* 1. A country merehant owes in Boston, 8200 due in 2 months, and S200 dne in 6 months, each w ithouc interest ; at what time could he pay both debts, that nekher party may lose ? Solution. — He may keep the first ms kmg after it is dne as he pa3r8 the last befo^ it is due. Ans. 4 months. The method of finding the time when several debts, due at difierent times without interest, should be paid at once, is called Equation of Payments. The time of payment thus found, is called the mean time. 2. A man owes S106, due in one year, and $106, due in 3 years, without interest ; in what time shall he pay both at once? Solution. — At the end of 2 3rears. But this, which is the common method, is a gain, in this example, of S'36 to the debtor. He keeps the first debt a year after it is due, and Ihns gains (at 6 per cent.) the inter- est on $106 for a year, or $6*36. He pays the whole of the second debt a year before due, when he should pay only such a sum as would amount to $106 tn a year, or $100. Thus he loses $6^ on the second debt, while he has gained S6'36 on the first. The error, which results from consid- ering the interest and discount on the same sum for die same time equal, is not usually regarded in business. , IT 181* To find a rule for the eqiuxtion of payments. 1. Borrowed of a friend $6'00, for 4 months ; afterwards 1 lent him $1, to keep long enough to balance the use of the money borrowed ; bow long must this be ? Solution. — He should keep $1 six times as long, or, Atis. 24 months. 2. In how long time will $8 be worth as much as $40 for 1 month ? Solution. — Every $8 in me $40 will be worth as much in 1 month, as the first sum, $8, is worth in that time. Then as many times as $8 are contained in $40, so many months the $8 will require to be worth as much as the $40 for 1 month. Ans. 5 months. 3. I have 3 notes against a man : 1 of $12, due in 3 Questions* — IT 180* On what prirciple is the time of paying at once the two debts, Ex. 1, detennined? W^M is understood by the mean time What is equatlypn of payments 1 What frror appears, Ex. 2, and why? 2(>* ^ EQ.UATION OP PAYMENTS. IT 181 months; 1 of $9, due in 5 months; 1 of $6, due in 10 months, aU without interest ; when should he pay the whole at once ? SoLUTioir. — $12 for 3 moiuhs, is the same as $S6 Utl month. S9 « 5 << ** $45 *< $6 " 10 « « $60 " He has $2'. long enough to balance $141 for 1 month. Every $27 m $141 will be worth as much in 1 month as the first $27. Then as many times as $27 is contained in $141, so long he can keep tlie $27. 141 -1-27 = 5 mo. 6-}- days, Ans Hence, To find the mean time of several payments. Multiply each sum by its time of payment, and divide the sum of tne products by the sum of the payments. BXAMPIiES FOR PRACTICBL 4. A western merchant owes in New York city $300, due in 6 months ; $325^50, due in 3 months, and $413^37, due in 2 months ; l)ut he finds that it will be more convenient to make payment at one time ; in what time will this be ? Ans. 2*984 months =3 months 29 ~\- days. 5. I owe several debts, due in difierent times, without interest, namely, $309*50, in 8 months ; $161, in 5 months and 18 days, and $63*25, in 10 months and 11 days; what shall I pay now te cancel the whole, the rate being 6 per cent. ? Note 1. First find the mean time^ then the present worth, IT 164. The fidction of a day will not be regarded m business operations. Ans. $514*375+. 6. A merchant has owing him $300, to be paid as follows : $60 » m 2 months, $100 in 5 months, and the rest in 8 months : and it is agreed to make one payment of the whole ; fa what time ought that payment to bel Ans. 6 months. 7. A owes B $136, to be paid in 10 months ; $96, to be paid in 7 months ; and $260, to be paid in 4 months ; what is the equated time for the payment of the whole ? Ans. 6 months 7 days -f- 8. A owes B $600, of which $200 is to be paid at the present time, 200 in 4 months, and 200 in 8 months ; what is the equated time for the payment of the whole ? Ans. 4 months. 9. A owes a $300, to be paid as follows : | in 3 months, 4 in 4 months, and the rest in 6 months ; what is the equated time ? Ans. 4j| months. Note 2. Sometimes retailers sell on 6 months' credit, without interest. But as it would be difficult to settle each item of a long account just 6 months from the time of purchase, all the charges for a jear are settled at its dose. Presuming that the purchases each month are uniform, this gives 6 months as Questions. — V 1181« Give the solution of Bx. Z. Give the solution of •he 3d example. Rul% Give the substance of iie notes. N UieBMiuitiiMof asatUc^neat Shoiil4 a settlement be niadt at the en< of 8 months, the mean time wonld be 4 months ; at the end of 6 months, ^e mean time wonld be S months. RATIO. 5T 199« How many tknes is 4 ccmtiuned in 8 ? Afu. 2 times. Te find how many iknes one number is contained in an- other, is to find the ratio between the numben, which we do by dividing one of the numbers by the other. But without performing the division we rfisy express it. First, by the sign of division : thus, 8 -r- 4. Second, by a Ime without dots, writing the divi- dend in place of the upper, and the divisor in place of t}ie low^ dot : thus, f . Third^ by dots without a line : thus, 8:4. The last is the usual method of expressing ratio, when the quotient in division r^eives this name. Hence, Ratio is the quotient expressing how many times one num- ber is contained in anotlier, or how many times one quantity is contained in pother of the same kind. Note. — Ratio can only enst betipeen quantities qfthe $ame kind, since the dividend and divisor most be of the same kind. It would be absurd to inquire how many times 3 bushels of rye are contained in 12 lbs. of butter. (See IT 31.) Biit a ratio can exist between numbers of different denomina- tions when they can be reduced to the same denomination ; thus, we can determine how many times 8 quarts are contained in 6 gallons, when we reduce the quarts to gallons, or the gallons to quarts. A ratio requires two numbers, each of which is called a term of the ratio, and together they are oaUed a couplet. The first term, whi<^h is the dividend, is called the antecedent ; the second, or divisor, is called the consequent. Hence it follows that multiplving the antecedent or dividing the conse- ?uent multiplies the ratio, (f^^,) dividing the antecedent or multiplving he consequent, divides the ratio, (IT 57,) and multiplying or dividing both antecedent and consequent by the same number does not alter the ratio, (IT 68.) Inverted and Direct Ratios. f[ tSfS. In the ratio 8 ; 4, the first term is divided hy Questions. — IT 183. What is it to find the ratio of numbers, and how done ? Describe the three ways of expressing the division. What is said of the last way ? Define ratio. How does ttie note limit ratio ? Why ? Illustrate. How many numbers tre required, and how many difiier^t Jl^anies do they receive ? Apply IT 56;— iT 57; — 1(58. 236 PROPORTION. T 184, 18S. the second, and the ratio is said to be direct. But sometimes the second is divided by the first, and the raLo is then said to be inverse, since it is equivalent to- inverting the terms, and writing the expression 4:8. The latter is ;ilso called a re- ciprocal ratio, since |, the quotient of 4 : 8, is the reciprocal of 2, the quotient of 8 : 4, (IT 55.) Hence, Direct ratio is the quotient of the antecedent divided by the consequent; and Inverse or reciprocal ratio is the quotient of the consequent divided by the antecedent. Compound Ratio. V 184» We have seen, IT 79, that a compound fraction consists of several simple fractions, to be multiplied together. Thus, the numerators of the compound fraction j of A^, are to be multiplied together for a new numerator, and the de- nominators for a new denominator. But since the terms of a ratio may be vnritten fractionally, we may call the expression a compound ratio. Hence, A compound ratio consists of several simple ratios to be multiplied together, which is done by multiplying together the antecedents, and also the consequents. PROPORTION. . IT tSS. 1. If 13 yards of cloth cost $18, what will 4 yards cost ? Solution. — As 4 yards are | of 12 yards, they will cost | of $18, or S6. The 12 yards contc.in 4 yards as many times as $18 contain $6; that is, 12 -a- 4 =s 18 -;- 6 ; or fractionally, -J^ = ■^. We have here two ratios, which are equal. But as the sign of divi- sion is written to express ratio without a line, 4 dots may be written to express the equality of the ratios. The four dots are used instead of the lines usually emploved. The expression then becomes 12 : 4 : : 18 : 6, equivalent to -^ ==-y*-. Such an expression is called a proportion, and is read, 12 divided by 4 equals 18 divided by 6 j or more commonly, 12 is to 4 as 18 is to 6. Hence, Qaestions* — IT 183* How does inverse, dilTer from direct ratio? De- fine each. What other name has inverse ratio? Why ? IT 184* Whence arises compound ratio? Defiae it. Give an example Write it in the common form. Reduce it to a simp.b ratio» T 186. PROPORTioiv. 237 Proportion is the combination of two equal ratios. The first and last terms are called the extremes, the second and third, the means. The two antecedents are called corre- sponding terms, as are also the two consequents, since these terms have always a certain reference to each other. The third term, $18, of this proportion, expresses the value of 12 yards, the first term ; and the fourth term, $6, expresses the value of 4 yards, the second term. NoTK. — A proportion requires fenr terms, two antecedents and two conse- joeots. Three numbers may form a proportion if one is used in both ratios ; Uuis, with the numbers 13, 6, and 3, we have the proportion, 13:6::6:3. Rule of Three. V 186* When, as in the ahove example, the first three terms are given to find a fourth, we may find it by taking such a part of the third term as the second is of the first ; or by a method called the Rule of Three, on the following prin- ciples. Take the proportion, 12 : 4 : : IS : 6, fractionally expressed, V='V^' As the fractions are equa], if we reduce them to the common denominator, 24, by the rule, IF 70, the numerators will be equal, and the fractions will become, j-| = j^. The first numerator, 72, it may be seen, is the product of the extremes, and the second numerator, 72, is the product of the means, of the proportion. Hence, The product of tJie extremes of a proportion is equal to the product of the means. Take the first throe terms to find the fourth. yds. yds. doU. doll. Solution. — "We multiply together 12t4jjl8:«»»» 4 and 18, the means, which gives the 4 product of the extremes, of which one — extreme, 12, is given, and we divide 12 ) 72 the product by the given factor to get — the other extreme, or fourth term. 6 doUars, Ans. Qnestions* — IT 185* How is the price of the 4 yards, Ex. 1, found? What two ratios are formed 7 By what sign is their e<iuality expressed, and instead of what is it used 7 Give the two ways of reading a proportion. De- fine proportion. What are its terms? What are extremes 7 What are means 7 What are corresponding terms, and whv 7 How many terms, and Itow many numbers, aie requirt i m proportion 1 Illustrate. FROPORTIQN. If 187, 168. Nora. — TMs oporation is strictly tiaalytiei for had the |rioe of 1 yvd been 18 dollars, we should have multiplied it by 4 to get the price of 4 ytuds. But as it is 12 yards, which cost 18 dollars, the product of 18 by 4 is 12 tim^ too krge, and most be divided by 12. V 187« To write down the three given numbers. S. At $90 for 15 barrels of flour; how many barrels can be bought for $30 ? doll. doU. bar. bar, Somnoir. — We have the tenns of one 90 * 30 t : 15 : •• •• ratio, f90 and S30, being of the same 30 kind. For the same reason, 15 barrds and the required number of barrels will 910)4510 form another ratio. We place the ante- cedent of the second ratio, 15 barrels, for 5 bar.f Ans, the third term, and its correspondent, $90 for the first, and $30 for the second, that its correspondent may be the fourth term. We then multiply and divide as above, and get the Ans.f 5 barrels. V 188« To invert both ratios. In the proportion, 90 : 30 : : 15 : 5, the quotient of 90 -4- 30 is 3, and that of 15 -f- 5 is 3. Inverting the terms df each couplet, we still have the proportion, 30 : 90 : : 5 : 15, for each ratio is now J, the reciprocal of the former ratio, (IF 183,) and consequently the two ratios are equal. This inversioa often virtually occurs in operations ; thus, 3. At $30 for 5 barrels of flour, how many barrels can be bought for $90 ? doll. doU. bar. bar, 90 : 30 :: .... 5 Somriow.—Writmg the first ratio, 90 .-so. By inversion, ^ above, 5 barrels must be the fourth terniy 30 "90 • • 5 • •••• ^ ^^ corresponds to $30, the second ter^i. • s** * B^^ i^ is convenient always to regard the 2 fourth as the nnkn> *vn term, and it will be o I A \ 4c I A come so b^ inverting both ratios, when the o I vi ;^ I u operation is the same as before. 15 bar., Aws. ^'«- ^^ ^"^^^^ Qaestions* ^ f 186. What is the object of the rtde of three 7 Com- pare the two methods of expressing a pro])ortion, and show what r^iults from reducing the fractions to a common denominator. How is the first numerator obtained 7 — the second 7 What important conclusion is derived 7 ^How ap- plied to finding the fourth term 7 How explained analytically 7 V 18T. What must be made the third term? Why 7 What must be made the first, and why 7 What the second 7 Why 7 88. Why oan both ratios be inverted? What is the object ia so dflS^ 1 189, 190. PROPORT ON. 239 1" 180* T> invert tme ratio. 4. If 3 men will boikl a wall in 10 days, in how long time will 6 men build it ? men.mtn. days, days, SottrriDH. — Writing the corre- 3 : 6:: 10: •••• Bpondmg terms as already described Invertmg the first ratio, we have, ^^ ^^"^ \ 5* ^ *«. ^^ to the required . 0..1A number of days, smc« 10 days IS the 6 : J : : lU time required by 3 men, and the un- 3 known term the time reqnired by 6 men. But since twice the number g \ QA of men will build the wall in half ' the time, 3 men are sneh a part ot (are contained in^ 6 men as the re 5 day$i Am, quired number or days are a part of 10 days. The first, then, is an in- verse ratio, the consequent being divided by the antecedent. But if we invert its terms, the division will be as usual, and the operation as al- leady described. * * KoTS. — Id the third example, more money would buy more flour ; in the second, less money wogld buy less flour. In such examples, each antecedent is divid^ by its consequent, or if one ratio is inverted, the other is also. The proportion is then called direct. But when, as in the fourth example, more requires less, (more days, less time,) or, as may be the case, less requires more, one ratio is inverse, while the other is direct. The proportion is then called inverse. Hence, Direct proportion is the (Combination of two direct, or two inverse ratios. Inverse proportion is the combination of a direct and an inverse ratio. But if one of the ratios is inverted, it may be treated as a direct proportion. IT 190» Hence, to find the fourth term of a proportion when three terms are given, we have thefoUcwing I. Make that one of the three numbers the third term -which is of the same kind as the aftswer sought, reducing the other two, if necessary, to the same denomination, that the terms of each ratio may be of the same kind. II. Write that one of the remaining two for the first term -which corresponds to the third, and the other for the second term. Questions. — IT 189* How are the numbers, Ex. 4, to be written accord- ing to the rule for the correspondin|[ terms ? What happens from this man- ner of writing them, and way 7 What then is done 7 How do the 'second mod third difi^ fix)m the faurth example 7 What is direct proportion 7 What imverae? I S40 PROPORTION. T 19L III. If the question involves the inverse proportion, more requiring less, or less requiring more, invert the first ratio. IV. Multiply together the second and third terms, or two means, to get the product of the extremes, which heing divided by the first term, or known extreme, the quotient will be the other extreme, or fourth term sought, of the sam^ kind as the third term. NoTB. — If the third tenn is a componnd, or a mixed mnnber, it must be made of but one denomination. SXABfPIiES FOR PRACTICES. V 191* 1. If 6 horses consume 21 bushels of oats in 3 weeks, how many bushels will serve 20 horses the same time ? horaet, honet. oalt, oaU. 2 7 Note 1 . — Since in the operation there are 0* 20 * * ^Jt * ••*•• ^^"^ys two numbers to be multiplied to- • ^ • • ^ • • • • • (retlier, and their ]»oduct to be divided by a 7 tnird number, the process may frequently be .._ shortened by cancelation, as shown ; the o \ 1 Af\ factor, 3, bemg canceled in 21 and 6, and ^ / ^^" the remai{iinz rectors, 7 and 2, being used as multiplier and divisor. 70 bush., Arts. 2. The above question reversed. If 20 horses consame 70 bushels of oats in 3 weeks, how many bushels will serve 6 horses the same time? Ans. 21 bushels. 3. If 365 men consame 75 barrels of provisions in 9 months, how much will 500 men consume in the same time t 73 15 jf0^:5OO::ar^: . ^715. 102^1- barrels. 4. If 500 men consume 102^^ barrels of provisions in 9 months, how much will 365 men consumer in the same time? . 100 73 $00 : i$$ : : 102ff : •• •• And reducing the third tern, to an improper fraction, we have, 100 ' Tti 2 : —=77- • • • • • Now, since the denominator is a divisor, (^ 64,) we cancel 73, and have, 100 : 1 : : 7500 : •• ••. Arts, 75 barrels. 5. If the moon move 13^ lO' 35'^ in 1 day, in what time does it perform one revolution? Arts. 27 days 7 h. 43 m. 6 s. +. Questions. — IT 190* What number is made the third term, and why 7 How are the other two numbers arranged 7 When is the first ratio inverted 1 For what is the multiplication 7 — the division 7 When is a r^ucUon needed 7 Repeat the whole rule. tm. ' PROPORTION. 211 6. Hft person, whose tent is $145, "pay ttQ*63 psnsh taxes, how much shoaid « person pay whose rent is $378 ? Atis. $32^025. 7. If I hay 7 lbs. of sugar for 75 cents, how many pounds can I buy for $6? Ans 56 lbs. Note 2. — Every example in proportion may be performed on general prin- ciples, IT 134. ThuSf in the last example, we may divide t*76,tne price of 7 U>s., by 7, and we shall have the price of 1 lb., and then divide SG) the price of the required number of lbs., by the price of 1 lb. now found, and it will give us the number of lbs. 8. If I give $6 for the use of $100 for 13 Humths, what must I give for the use of $357'82 the same time? Ans. $21'4694-. On general principles. Divide $6 by 100 to get the gain on $1, which multiply by the number of dollars, to get the gain on the num- ber of dollars. Note 3. — Let the pupil be required to perform all the subsequent examples both by proportion and on general principles, or analysis. 9. If a staff 5 ft. 8 in. in length, cast a shadow of 6 feet, how high is that steeple whose shadow measures 153 feet? Ans. 144^ feet. 10. If a family of 10 persons use 3 bushels of malt in a month, how many bushels will serve them when there are 30 in the family? Ans. 9 bushels. By Analysis. If 10 persons use 3 bushels, 1 person will use i\j of 3 bushels, or -]% of a bushel. And if 1 person use ^, 30 per- sons will use 30 times -j^= f-^=:9 bushels. Note. 4 — The seven following examples iavolve the inverse proportion. 11. There was a certain building raised in 8 months by 120 work- men ; but the same being demolished, it is required to be built in 3 months ; I demand how many men must be employed about it. ^715. 480 men. 12. There is a cistern having a pipe which will empty it in 10 hours ; how many pipes of the same capacity will empty it in 24 min- utes? Ans. 25 pipes. « 13. A garrison of 1200 men has provisions for 9 months, at the rate of 14 oz. per day ; how long will the provisions last, at the same allowance, if the garrison be reinforced by 400 men? Ans. 6| months. ;^ 14. If a piece of land 40 rods in length and 4 in breadth, make an acre, how wide must it be when it is but 25 rods long? Ans. 6§ rods. 15. If a man perform a journey in 15 days, when the days are 13 hours long, in how many wm he do it when the days are but 10 hours tong? -4.715. 18 days. 16. If a field will feed 6 cows 91 days, how long will it feed 21 cows? Ans. 26 days. Questions, — IT 191. How is the operation in projxHrtion shortened? How may every example in proportion be performed without stating iti What is tne method described, IT 134, for examples requiring several opera tlcns? 21 242 PROPORTION. ' T 192. f 17. Lent a friend 392 dollarB for 6 months; some lime after, he lent me 806 dollars ; how long may I keep it to balance the favor % Ans, 2 months 5 4~ days. 18. If 7 lbs. of sugar cost | of a dollar, what cost 12 lbs. ? yf^y-: Ans.tif. 19. If 6i yds. of cloth cost $3, what cost 9| yds! ATM. $4*269+. 20. If 2 oz. of silver cost $2*24, what costs | oz. ! Arts. $0*84 21." If ^ oz. cost $1^, what costs 1 oz.? Ans. $1'283. 22. Iff yd. cost $J, what will 40 J yds. cost? Ans. $59*062-}-. 23. A merchant, owning f of a vessel, sold | of his share foi $957 ; what was the vessel worth? Ans. $1794*375. > 24. If 12 acres 3 roods produce 78 quarters 3 pks of wheat, how much will 35 acres 1 rood 20 poles produce? / ^ '-^ T-"'^ i ' '^ y ) 7 V p^Ans. 216 qrs. 5 bush. 1 pk. 4 qts. - 25. A cistern has 4 pipes which will fill it, respectively, in l6, 20, 40, and 80 minutes ; in what timp will all four, running together, fiU it ? Ans. 5^ minutes. 26. At $33 for 6 barrels of flour, what must be paid for 178 bar- rels? Ans. $979. 27. If 2*5 lbs. of tobacco cost 75 cents, how much will 185 lbs. cost? Ans. $55*50. 28. What is the value of * 15 of a hogshead of lime, at $2*39 per hhd.? Ans. $0'3585. 29. If * 15 of a hhd. of Hme cost $0*3585, what is it per hhd . ? Ans. $2*39. 30. A bankrupt owes $972, and his property, amounting to $607*50, is distributed among his creditors ; what does one receive whose demand is $1 1 J ? Ans. $7*083 -|-. 31. When wheat is worth $*93 per bushel, a 3 cent loaf weighs 12 oz. ; what ipust it weigh when wheat is $1'24 per bushel? Ans. 9 oz. 32. A company of 16 men are on an allowance of 6 oz. of bread a day ; what will be their daily allowance for 28 days, if they receive an addition of 224 lbs. ? , , Ans. 14 oz. Compound Proportion. IT 199» 1. If a man travel 240 miles m 8 days of 12 hours long, how far will he travel in 6 days of 10 hours long, trav- eling at the same rate ? SoLxrrioif — The relation of the number of miles, 240, which must be made the third term, to the required distance, depends on two circum stances, the number, and the length of the days. 1 1212. raopoETioN. 843 first, coDsid*xing the nnmbor of days, without regard to taeir leogth, westiall have, 8 : 6 :: 240 : • • -^" - ^ • • • We find that he will travel 180 miles in 6 da3r8 of the same leiigdi as the 8 days. Second, considering the length of the days, without regard to their nomber, we shall have, hovra. kourt, rnUea, miUt, 12 : 10 :: 180 : .... He will travel 150 miles in 6 da3rs, 10 hoars long, vhen he travels 180 in the same number of days, 12 hours long. By the dependence of the dis|tance on the ratio 8 : 6, we get } of 240, and on the ratio 12 : 10, we get -f$ of this |-. But -{^ of f is a com- pound fraction, and consequeutly, 8 : 6 and 12 : 10, on which the dis- tance depends, constitute a compound ratio, which may be united with the given distance, as follows : _ ^ The compound may be Ccn.p.ratio. j J ! iJ } :: 240 ; .... g^wSiao^ Eeduced. 96 : 60 : : 240 : .. .. fration is the same as m a simple proportion Such a proportion is called a compound proportion. Hence, A compound proportion is the combination of a compound and a simple ratio, and exists when the rel/tion of the given quantity to the required quantity of the same kind depends on several circumstances. Note 1. — That the compound does not differ from the simple proportion, may be seen from the fact that, in the above reduced proportion, 96 is the number of hours in which 240 miles are traveled, and 60 the number ot hours in which the required distance is traveled. 2. If 264 men, in 5 days of 12 hours each, can dig a trench 240 yards long, 3 wide, and 2 deep, in how many days, of 9 hours long, vnll 24 men dig a trench, 420 yards long, 5 wide, and 3 deep ? Solution. — The number of days is required, and its relation to the given quantity, 5, depends on five circumstances, the number of men, the length of the days, the length, breadth, and depth of the trench. Placing 5 for the third term, we connect it with a compound ratio com posed of five simple ratios j thus, Questions*— IT 192. Why two operations to Ex. 1? Describe the first ; — the second. How does it appear that the relation of the distances depends upon a compound ratio 7 What is a compound proportion ? How is it seen that it does not differ in principle from the simple proportion 7 What reduction is made on the compound ratio 7 How is the operation of reducing shortened 7 Give the rule. * iM" SfiOPOKtWIt'. " ^ iTwerte, 3 4 Jnverte, i m 4 4 7 ' « Diret^, nm : ^0 Direct, $ : 5 Direct, 2 : 0J 11^483^ dayt. ••5 3x2 = 6. 11X7X5 = 385. Each simple ratio is from one circnmstance, regardmg the otlier eir camstances uniform. The first two ratios, it may be seen, are inverse since the less the number of men employed, or the shorter the days, the greater will be the number of da3rs. Reducing this compound ratio to a simple one, shortening: the process by cancdation, we hare the simple proportion, — 6: 385:: 5 By this proportion we get the ani^er In the 6 ) 1925 ^ifuoit manner as by any single proportion. 320f days. Arts, V 103* Hence, questions involving a compound propov' tion, may be performed by the following I. Having" made that number the third term which is df the same kind as the answer sought, we form £l ratio, either direct or inverse, as required, from the two remaining num- bers which are of the same kind, and place it for one couplet of the compound ratio. Form each two like remaining num^ bers into a ratio, and so on, till all are used. II. Having reduced the compound to a simple ratio, by multiplying together the antecedents and ^he consequents, shortening the process by cancelation, the deration will be the same as in a simple proportion. BXAMPLOBS FOR PRACTICES. 1. If 6 men build a wall 20 ft. long, 6 ft. high, and 4 ft. thiek, in 16 days, in what lime will 24 men build one 200 ft. long, 8 ft. -high, and 6 ft. thick? Ans. 80 days. 2. If the freight of 9 hhds. of sugar, each weighing 1200 lbs., 20 miles, cost $ 16, what must be paid for the freight of 50 tierces, each weighing 250 lbs., 100, miles. Questions* — IT 193« What is the rule for compound proportion? How many, and what circumstances in Ex. 2 7 How ^se may the examples be performed? of caskS) the size of the casks, and toe distance. Ans. $92*69 +. 3. If 56 lbs. of bread be sufficient for 7 men 14 days, how mnch bread will serve 21 men 3 days? Arts, 36 lbs. The same by analysis. If 7 men consume 56 lbs., 1 man will con- sume -f of 56 = 8 lbs. in 14 days, and -j^ of 8 lbs., = A of a lb., m 1 day. 21 men will consume 21 times what 1 man wiU consume, that is, 21 times '^=^f^= 12 lbs. in 1 day, and 3 times 12 lbs = 36 lbs. in 3 days. NoTS 2. — Havinff wroo^t the fidlowing examples by fwp9ttum, let llie pupil be required to do the same by analysis. 4. If 4 reapers receire $11*64 fbt 3 days' woik, how many men may be hired 16 days for $103*04 1 Ans, 7 men. 5. If 7 ox. 8 drs. of bread be bouglit for $*06 when wheat is $'76 per bushel, what weight of it may be bought for $*18 when wheat is $*90 per bushel? Ans, 1 H). 3 oz. 6. tf $100 gain $6 in 1 year, what will $400 gain in 9 months! Note 3. — This and the three following examples ledprocally prove each other. 7. If$it)0gfaiii$6iniye», in What time will $40a gain $lSt 8. If $400 gain $18 in 9 months, what is the rate per cent, per annum! 9. What principal, at 6 per cent, per annum, will gain $18 in 9 months! 10. A usurer put out $75 at interest, and at the end of 8 months, received, for principal and interest, $79 ; I demand at what rate per cent, he received interest. Ans. 8 per cent. IT tWk. Review of PropaxtUm. QnestfoBi^ — What is ratio? Between what qnantities does it exist? What is inverse ratio? —compound ratio? What is propor- tion? rule of three? How is it shown that the product of the ex- tremes equals the product of the means ? What inversions of the ratios take place ? How is it shown that the operation of the rule of three depends on analysis? What is inverse proportion? —direct propor tinn ? — compound proportion ? BXBILGISIXU 1. If 1 buy 76 yds. of cloth for $113*17, what does it cost per eU English? ^ :y, . ^ , . ; ^n5. $1*861 -^. 2. Bought 4 pieces of Holland, each containing 24 ells English, for $96 ; how much was that per yard ? iln*.$0*80. al. A garrison had provision for 8 months, at the rate of 15 OMieet 21* 346 ALLIOATION. IT 195 to each person per day ; how much must be allowed per day, in order that the provision may last 9^ months! Ans, 12|-^ oz. 4. How much land, at $2'50 per acre, must be given in exchange for 360 acres, at $3^75 per acre? Ans. 540 acres. 5. Borrowed 185 quarters of com when the price was 193. ; ho\« much must I pay when the price is 17s. *4d. ? Ans. 202^^ qrs. 6. A person, owning f of a coal mine, sells | of his share for JC171 ; what is the whole mine worth 1 Ans. jC380. 7. If ^ of a gallon cost f of a dollar, what costs f of a tun 1 Ans. $140. 8. At J^l^ per cwt., what cost Z\ lbs. 1 Ans. lOfd. 9. If 4i cwt. can be carried 36 miles for 35 shillings, how many pounds can be carried 20 miles for the same money! Ans. 907| lbs. 10. If the sun appears to move from east to west 360 degrees in 24 hours, how much is that in each hour! in each minute! — in each second ! Ans. to the last, 15'^ of a deg. 11. If a family of 9 persons spend $450 in 5 months, how much would be sufficient to maintain them 8 months, if 5 persons more were added to the family! Ans. $1120. ALLIGATION— Medial. IT 195. 1. A farmer mixed 8 bushels of com, worth 60 cents per bushel, 4 bushels of rye, worth 80 cents per bushel, and 4 bushels of oats, worth 40 cents per bushel ; what was a bushel of the mixture worth ? When several simples of different values are to be mixed, the process of finding the average price, is called Alligation Medial, The average price is called the mean price, OPERATION. Solution. — Multiply 8 bushels, at peO ; 60 X 8 =$4*80. 1,^^ f;60, ;[|;^P^f ^^f, j" 4 m^hels, at $'80 ; 80 X 4= $3^20. J^^^itl'trpn^ce^^^^ . j4 bushels, at $'40 ; 40 X 4 = $1*60. bushels. In like manner i/» J r T t iikr^,nn WC find thC priCC Of ihc 16 bushels are worth $9*60. ^ye^ and the oats in the 1 bushel is taorth $'60. mixture, and addmg to- gether the prices of the corn, the rye, and the oats, we have #9'60, the price of 8 -f- 4 -j- ^ ■* 1^ bushels, which are contained in the whole mixture, and dividing the price of 16 bushels by 16^ we get the price of 1 bushel. Ans. 60 cents. Hence, the Qoestiont. — IF 195. What is alligation medial 7 --mean rate 1 What Ib the rule 7 To what does it apply 7 ! f 196. ALUGATION. 247 Find the prices of the several simples, and add them to- gether for the price of the whole compound, which divide by Sie number of pounds, bushels, &c., to get the price of 1 pound or bushel. NoTB. — The prineipies of the rale are applicable to many examples not embraced by the aboTe definition of Alligation Medial. KXAMPLJBS FOR PRACTICE* 2. A grocer mixed 5 Ibe. of tagar worth 10 cents per lb., 8 lbs. worth 12 cents, 20 lbs. worth 14 cents ; what is a pound of the mix- ture worth ? Ans, $* 12f ^. 3. A goldsmith melted together 3 ounces of gold 20 carats fine, and 5 ounces 22 carats fine ; what is the fineness of the mixture I Ans, 21^ carats. " 4^ A grocer puts 6 gallons of water into a cask containing 40 gal* Ions of rum, worth 42 cents per gallon ; what is a gallon of the mix- ture worth ? Ans. 36 j^ cents. 5. On a certain day the mercury was observed to stand in the ther- f mometer as follows : 5 hours of the day it stood at 64 degrees ; 4 1^ hours, at 70 degrees : 2 hours, at 75 degrees ; and 3 hours at 73 de- I N . grees ; what was the mean temperature of that dayl Ans, 69^ degrees. 6. A farm contains 16 acres of land worth $90 per acre, 22 acres worth $75, 18 acres worth $64, 10 acres worth $55, 30 acres worth $36, and 42 acres worth $25 per acre ; what is the average value of » the farm per acre? Ans. $50*16 nearly per acre. 7. A dairyman has 20 cows, 3 of wliich are worth $35 each, 4 are worth $30, 6 are worth $24, 4 are worth $20, 2 are worth $18 each, and 1 is worth $13 ; what is the average value? Ans, $24^90. Alligation Alternate. IT 196. 1. A fanner has 1 bushel of com, worth 50 cents. How many bushels of oats, worth 40 cents, must be put with it, to make the mixture worth 42 cents ? SoiuTTON. — The 1 bushel of com is worth 8 cents more than the pricfc of the mixture, and as each bushel of oats is worth. 2 cents less, he must take 4 bushels of oats Ans. When the price of several simples, (corn and oats,) and the price of a mixture to be formed from them are given, the method of finding the quantity of each simple is cdled AUU gation AUemate. ALLIGATION. If 196. 2. How many bushels of oats must the fanner take to mix with 2 bushels of corn, prices the same as above ? Solution. — Evidently, twice as many as were required for 1 bushel, of, Ans. 8 bushels. Note 1. — We see that 2, the number of bushels of com, equals me numbei of cents that the oats are worth less . than the mixture, and S, Ihe number of bushels of oats, equals the number of cents that the corn is worth m&re than the mixture. Then, if the three prices had been gflTen, we might have taken 2, the differ- ence between the price of the oats and of the mixture for the number of bushels of com, and 8, Ihe difference between the priojB of the com and of the nilxtuile, for the number of bushels of oats, and we should hate soch a mixture. That is, toe take tke difference between the price of each sim- ple and of ihe mixture for the number of hiishels of the other simple, NoTB 2. — By this process, the sum of the excesses, found by multiplying 8 cents, the excess of 1 Dushel of com, by 2, the number of bosnels, equals the sum of the deficiencies, found by multiplying 2, the deficiency of I oushel of oats, by 8, the number of bushels. Or, 8 X 2 =^ 2 X 3) s^Josa the fieictors are the same in each. 3. A merchant has two kinds of sugar, worth 6 cents and 17 cents per pound, of which he would make a mixture worth 10 cents ; how much of each must he take ? OPEBATION. Solution. -a- Take the Ti • r •-x. i/\ J. { 6n71bs. difference between 10 Fnce Of mixture, 10 cents, j j,^ J ^ ^^^ and 17 for the nunibcr ^ * of lbs. at 6 cents, and the difference between 6 and 10 for th^ number of lbs. at 17 cents. The sum of the deficiencies, 7 times 4 cents, equals the sum of the excesses, 4 times 7 cents. Note 3. — This gives a mixture of 11 pounds, and if 3 times, 5 times, one half, or any other proportion of the mixture be required, the same proportion of each simple must evidently be taken. The same would be done if 3 times, 6 times, &c., one simple were given. 4. A merchant has two kinds of sugar, worth 8 cents and 13 cents ; how much of each must he take for a mixture worth 10 cents per pound ? OPERATION. &n 3 lbs. at 8 cents. cents. Price of mklure, 10 cents, j jf] % j^; ^J J Note 4. — The price of the mixture in the last two examples is the same. 6. A merchant has foui kinds of sugar, lyorth 6, 8, 13, and 17 cents per pound ; how much of each must he take for a mixture worth 10 cents ? T 107. ALLIMttom 7^ 6 with 17, to show th«t I (^ o sugar at those prices are Price of mixture, 10 cents. 4 ,XJ t' to be mixed, we have 7 * lo-" 2. lbs. at 6 cents and 4 lbs. 17— -' 4. at 17 cents, forming a raiztore of 11 lb*« wckth 10 cenrs; Ib likfr mamici!, w« have a mixtnie of 5 Iba. irarth 10 oeata^ lO«amQ0Otift|^8aBdia. And U-f^^-l^Uto-iAalL NoTB 6. — We may see in this and all examples, that the sum of the de&- cSendes equals Ae sum of the excesses. 6. A merchant iias 3 kinds of sugar, worth 6^ 8^ and 15 cento ; ho^ nrach of each must he take for a mixture worth 11 cc^ts per pound ? OPERATION. SOLTTMeW.— / g — , ^ We have not Price of mixture, 11 cents. ) Sn i 4. • ^^oTforme? ( 15-^ 6 -f- 3 s= 8. example, but 4 lbs. at 6 cents, may be mixed with 5 lbs. at 15 cents, and 4 lbs. at 8 cents, with 3 lbs. at 15 cents. The 15 is connected with both, because we take two por- tions at this price, 5 lbs. to mix with 4 lbs. at 6 cents, and 3 lbs. to mix with 4 lbs. at 8 cents. We have, then, 8 lbs. at 15 cents. IT lV7-« From the examples explained, We have the fol lowing I. Write the prices of the simples directly under each other, h^^ning with the least, and the price of the mixture at the left hand. II. Connect each price less than the mean price with' one or more greater, and each price greater with one or more that is less; III. Write the difference between the price of the mixture and of each simple opposite to the price or prices with which it is connected ; the number or numbers opposite to the price of each, simple will express its quantity in the mixture. IV. If any fraction or multiple, as one third^or ^ree times one of the simples is to be taken, the same fraction or multi** Questions* -> IT 196* Why 4 bushels of oats to 1 of corn. Ex. l and 2 7 Wmtt equalities appear from 'Ea..2l How could we- have made such a mix* ture as we have^ it the prices onhr had been known 7 Give the general method of forming sucn mixtures. What are equal, as explained in note 2 7 Why 7 What must be done if a greater or less quantity of the mixture be reqirir€d7-* of efte^imple 7 Why are mmibefs cenn^eted 7 What is done, in Ex. 6, when Jiere are nni two pairs 7 What is alligation alternate 7 iri97. What is the rule 7 25a ALUOATlon. T 197 pie of each of the other simples will be required. Or, if any iraction or multiple of the whole compound be required, the same firaction or multiple of each simple must be taken* BXAMPLES FOR PRACTICE. 1. What proportions of sugar, at 8 cents, 10 cents, and 14 c^its per pound, mil compose a mixture worth 12 cents pec pound t Atis. In the i^oportion of 2 lbs. at 8 and 10 cents to 6 lbs. at 14 cents. 2. A grocer has sugars, worth 7 cents, 9 cents, and 12 cents per lb., which he would mix so as to form a compound worth 10 cents per pound ; what must be the proportions of each kind ! Ans. 2 lbs. of the first and second, to 4 lbs. of the third kind. 3. If he use 1 lb. of the first kind, how much must he take of the others? if 4 lbs., what? if 6 lbs., what? if 10 lbs., what? if 20 lbs., what? ^715. to the last, 20 lbs. of the second, and 40 of the third. 4. A merchant has spices at 16d., 20d., and 32d. per pound ; he would mix 5 pounds of the first sort with the others, so as to form a compound worth 24d. per pound ; how much of each sort must he use ? Ans, 5 lbs. of the second, and 7^ lbs. of the third. 5. How many gallons of water, of no value, must be mixed with SO gallons of rum, worth 80 cents per gallon, to reduce its value to 70 cents per gallon ? Ans. 8f gallons. 6. A man would mix 4 bushels of wheat, at $1^50 per bushel, with rye at $1*16, corn at $'75, and barley at $'50, so as to sell the mix- ture at $'84 per bushel ; how much of each must he use ? 7. A goldsmith would mix gold 17 carats fine with some 19, 21, and 24 carats fine, so that the compound may be 22 carats fine ; what proportions of each must he use ? Ans. 2 of the first 3 sorts to 9 of the last. 8. If he use 1 oz. of the first kind, how much must he use of the others ? What would be the quantity of the compound ? Ans, to the last, 7^ ounces. 9. If he would have the whole compound consist of 15 oz., how much must he use of each kind ? if of 30 oz., how much of each kind ? if of 374 oz., how much ? Ans, to the last, 5 oz. of the first 3, and 22j| oz. of the last. 10 . A man would mix 100 pounds of sugar, some at 8 cents, some at 10 cents, and some at 14 cents per pound, so that the compound may be worth 12 cents per pound ; how much of each kind must he use' 20 lbs. at 8 cts. ) 20 lbs. at 10 cts. \Ans. 60 lbs. at 14 cts. ) 11. A grocer has currants at 4d., 6d., 9d., and lid. per lb., and he would make a mixture of 240 lbs., so that the mixture may be sold at 8d. per lb. ; how many pounds of each sort may he take? Ans, 72, 24, 48, and 96 lbs., or 48, 48, 72, 72, Sic. NoTB. — This question may have five diflerent answers. 1 T 198, 199. EXCHANQB. 251 EXCHANGE. IT 198* If a firmer, A, has com, and a manufacturer, Bj has cloth, each more than he needs himself, while he wants some of the article possessed by the other, an exchange will be made for mutual accommodation. But if B does not want A's com, while A still wants the cloth, the latter must find a third p:rson, if possible, who wants his corn, and can givt him something for it which B may want. This might be difficult, unless some article was settled on, which every one would take; then A might exchange his com for it, as B would part with his cloth for such an article, since, if every one would take it, he could procure with it whatever he might desire, though he did not want it himself. Such an article is called money. Gold and silver, contain mg great value in little space, are employed for money amonc civilized nations, and sometimes copper, to represent smal' values. This exchange between individuals is called trade, or commerce. Note 1. — Since gold and sHver, in their pure state, are too flexible for the purposes of a circulating medium, nine parts of pure gold and one part of sil- ver and copper, in et^ual quantities, are used by the U. S. government, for gold coins, nine parts of silver and one of copjper. for silver coins. The baser metal, m each instance, is called alloy. The English government uses only one ^tart of allov to eleven of the gold, and a little less alloy in silver coins. Hence, English coins are more valuable than ours of the same weight. Again, as coins are troublesome, and sometimes expensive to transport, and also suffer loss by wear, bank bills are much used for circulation, which, though valueless themselves, are readily taken as money, being payable in specie, on demand, at the bank^ which issued them. The coins are called specie^ in distinction from paper money, and together they form what is called the circulating me ■ dium, or currency. Note 2. — By ihe Jineneas of gold, is meant its purity, a twenty-fourth part of any quantity being called a carat. When, for example, there are two parts of alloy to twenty-two of pure gold, it is saia to be twenty-two carats fine. IT 199. Bank bills can be used in trade by individuals of the same country, but are not convenient in trade between those of different countries, since they would be removed too far from the place where payable. But the cost and risk of Questions. — IT 198« What trade are A and 6 supposed to make ? What will A do, if B docs. not want his com ? What is exchange 7 What is money? What are used for money, and why ? How much do coins want of l)eing pure gold and silver '? What is the value of English coins compared with ours, and why ? Why are bank bills used ? What do you understand to be the difier- ence between a bank bill now described, and a bank note, IT 174 1 What it specie? What is currency? What Is meant by fineness of gokl? ^by carat? — by carats fine? IlliistnUe. L 282 ' EXCHANGE. If 19d. transporting specie 3 considerable. If, for instance, Boston merc^hants purchase goods in Hamburgh to the amount of $2,050,000 a year, and the expense of transporting specie was 3 per cent., which is only a moderate allowance, this expense would be $61,500 to make payments for one year. If, on the other hand, Hamburgh merchants should purcnase $2,000,000 worth of goods, it would cost them $60,000 to make the pay- ments in specie. To reduce this expense, the following method has been de* vised. A B, a Boston merchant, has $10,000 due him in Hamburgh, from C D. He writes an order for the sum, and finds some one who owes $10,000 to another merchant in Hamburgh. He sells to him this order, and the purchaser sends it to his creditor, who goes to C D, A B's debtor, and receives the money. In this way, $2,000,000 of the Boston purchase might be balanced by the $2,000,000 purchased m Boston by Hamburgh merchants, and the Boston mer- chants would have to send only $50,000, the balance, thus reducing the expense of making payments between the two ports from $121,500 to $1500 ! Such an order is called a Bill of Exchange. Note 1 . — Lest a bill of exchange may be lost, or delayed, three copies are sent, by different conveyances, and when one is paid, the others are canceled. The form of one bill, to which the others agree, except iif the numbers of the bills,. is here given. Exchange for $10,000. Boston, Jan. 1, 1848. Three months after date, pay this, my first of exchange, (second and third of the same tenor and date not paid,) to the order of Flemming and Johnson, ten thousand dollars, value received, with, or without further advice from me. A B. CD, Merchant at Hamburgh. Note 2. If A 6 had to be at the expense of bringing from Hamburgh the Vpecie on his due, he could afford to sell his bill for less than $10,000, on ac- count of freight. This he would have to do when more was to be paid from Hambui^h to Boston than from Boston to Hamburgh, that is, if the balance of trade were in favor of Boston, as there would not be demand for all the Questions* — IT 199« "Vf hy are not bank bills convenient in exchange oetween different countries? What difficulty in paying with specie? Illus- trate by the example of trade between Boston and Hamburgh. How is A B ^ Hamburgh „_ . What care is taken in sending bills? Give a form. What would be the Ibrm of the second hill? — of the third ? When, and why, is exchange bekm 7 - above par ? Why do brokers deal in exchange ? ) T200. SXGHiNGB. 9A orders on nttnuur^ti) snd specio mnLa. hftTO to bo bTOUgnt OB Botto* Ex- change is then said to be baow ww. But if the mm:haser had lo pty ftr fireifhUi^ the SMiiiev he owes to Hambnrgh, be could afford to pijr mora than il 0,000 tor A B's bill, which be would have to do when the balance of trad*) was against Boston, tnat is, when Boston owed more to Hamburgh than it had owing, for then there would not be orders on Hambui^h sufficient to pay the debts, and some must send specie. Exchange is then said to be above pan , or at a premium. Note 3. — The broker is the medium between the seller and purchaser of bills, since the former might not know to whom he could sell, or the latter of whom he could buy. Brokers purchase bills, or take them to sell on commis- sion. The illustrations of the principles of exchange will be applied to exchange with England and France. Exchange with England. IT 900* The nominal ralue of the pound sterling is S4'44|>, consequently, a bill of exchange for £1000 is said to be worth S4444^44|. But by comparing the materials of the English sovereign, a gold piece representing a pound, and the eagle of our currency, the former is worth about J^4*86^. Sov- ereigns, however, are more or less worn by use, those dated as far kick as 1821 being worth no more than $4*80. It is pre- sumed that a quantity will average $4*84 each, and at this value they are taken in payment of duties. If, now, the nominal pound, $4*44|, be multiplied by *09, $4'444 X *09 ==*40, and the product be added to it, $4*44|+*40=$4*84| it will be changed to about its value, in the custom-house es- timations. If *09J be added to the nominal pound, it will become $4*86|, nearly its commercial value. When, then, sterling exchange is quoted at 9 or 9i per cent, advance, we must understand that bills sell for their par value. When above these rates, they are at a premium ; when below, at a discount. Noi'e 1. — In the following examples we shall consider H per cent, above the nominal, the par value. The pupil must not suppose, however, that 10 per cent, above the nominal, would be J j)er cent, above the real value of a bill. 1. A merchant sells a bill of exchange for JC5000 at its par value ; what does he receive? ^715. $24333'33i. 2. A merchant sold a bill of exchange for JETTOOO sterling, at 11 per cent, advance ; what did he receive more than its real value ? Ans, $466'66t. 3. A merchant sells a bill on London for £iWiO at 8 per cent, above its nominal value, instead of importing specie at an expense of % per cent. ; what does he savet An$, $133'66}. 22 254 EXCHANGE. H 20L 4. A, broker sold a bill of exchange for jC2000, on commission, at 10 per cent, above its nominal value, receiving a eommiscdon of ^ per cent, dh the real value, and 5 per cent, on what he obtained for the bill above its real value ; wha. wras his commission ? Ans. $ir95f. Note 2. — Though doUars and cents are the denominations of U. S. money, shillings and pence are much used in common calculations. But the dollar .las different values in different states, as expressed in sbillines ; thus, in New York, Ohio, and Michigan, it is 8 shillings ; in North Carolina, 10 shil- lings ; in New Jersey, Pennsylvania, DelaAvare, and Maryland, it is 7 shillings 6 pence ; in Geoi^ia and South Carolina it is 4 shillings 8 pence, while in the other states it is 6 shillings. The change of monev from one of these cur- rencies to the other is not now worthy of a formal discussion, as a metho<? will readily suggest itself to the practised arithmetician, and tne custom of using these denominations, it is hoped, will be speedily given up for the sim pier system of our federal currency. Thus, as 6 shillings in New Enelanc equal S shillings in New York, add one third of any number of shillings N. £ currency to the number, and we have the value expressed in shillings N. Y corrfQcy. Exchange with France. V 901. The unit of French money is the franc, the value of which is $*18f . In the quotations of French exchange, we have the number of francs that the dollar is rated at. As $1*00 7 is equal to ^Tqq- francs =5*37^^-1-, when a dollar is worth 6'37-^ francs it is at par. 1. A New York merchant sold a bill of exchange for $2500 on Havre, at 5*4 francs per dollar ; what did he obtain for it more than its value? Ans. $11. 2. A merchant bought a bill on Havre of $2800 at 5*31 francs per dollar ; what did he give less than its value? Ans. $34' 552. Questions* — IT 200* What is the nominal value of the English pound ? — the real value? What are sovereigns of 1821 worth? Why no more? What is the average value of sovereigns supposed to be, and where are they taken at this value? How is the pound changed from its nominal to its real value ? What is added to the nommal value in the examples ? What is said of 10 per cent. ? What denominations are still used in common calculations? What are the different values of the dollar in different states ? IT 201* Whot is the unit of French monev? — its value? What is the par value of the doUar, as expressed in francs 1 1202. BXCHANQE. 255 V 909. Value of Gold Coiiui. [According to the Laws passed by Congress^ May and June, 1834.] S. or Conn. d. cm. UNITED STATES. % Eaffle, coined beibra July 81, 1834, ' 10 66 6 Sbares in proportkm. FOREIGN GOLD. AUSTRIAN DOMINION& Souverein, 3 37 7 Double Ducat, 4 58 9 Hungarian, do., 8 29 6 BAVARIA. Carolin, 4 95 7 Max d*or, or MasimiUan, 3 31 8 Ducat. 2 27 6 BERNE. Ducat, double In pioportioiL 1 98 6 Pistole, 4 64 2 BRAZIL. Johannee, | In proportion, 17 6 4 Dobraon, 32 70 6 Dobra, 17 30 1 Moidora, | In proportion, 6 55 7 Crusade, 63 5 BRUNSWICK. Piaiole, douUa in proportion, 4 64 8 Ducat. 2 23 C0IX3GNE. Ducat, 2 26 7 COLOMBIA. Doiibloona, 16 53 6 DENMARK Ducat, Current, 1 81 2 Ducat, Specie, 2 26 7 Christian d'oc, 4 02 1 EAST LNDIES. Rupee. Bombay, 1818, 7 09 6 Rupee, Madras, 1818, 7 11 Pagoda. Star, 1 79 8 ENGLAND. Guinea, lialfin proportion, 6 07 5 Sovereign, do., 4 84 6 Seven SiiiUing Piece, 1 69 8 FRANCE. Double Louis, coined before 1786, 9 69 7 ]<ouis, do., 4 84 6 DouUeLouis, coined since 1786, 9 15 3 Ix>uis, do. do., 4 57 6 Double Napoleon, or 40 francs, 7 70 2 Napoleon, or 20 do., 3 85 1 FRANKFORT ON TOE MAIN. Ducat, 2 27 9 GENEVA Pistole, old, 3 98 5 Pistole, new, 3 44 4 GENOA. Sequin, 2 30 2 BAMBURG. Ducat, douUt la proportion, 2 87 9 Namm of Conra. d. e.in. HANOVER. Double Oeorgt d'or, slncle In proportion, 7 87 9 Ducat, 2 29 6 Gold Florin, doubto In propo^ tion, 1 67 HOLLAND. Double Ryder, 12 20 6 Ryder, 6 04 S Ducat, 2 27 ^ Ten Guilder pleco, 6 do. In pro- portion, 4 03 4 MALTA. Double Louis, 9 27 b Louis, 4 85 2 Demi Louis, 2 33 6 MEXICO. Doubloons, shares In proportion,* 15 63 6 MILAN. Sequin, 2 29 Doppia, or Pistole, 3 80 7 Forty Livre Piece, 1808, 7 74 2 NAPLES. Six Ducat Piece, 1783, 6 24 9 Two do., or Setiuin, 1762, 1 69 1 Three do., or Oncetta, 1818, 2 49 NETHERLANDS. Gold Lion, or Fourteen Florin Piece, 5 04 6 Ten Florin Piece, 1820, 4 01 9 PARMA. Quadruple Pistole, douUo In . proportion, 16 62 8 Pistole or Doppia, 1787, 4 19 4 do. do., 1796, 4 13 5 Maria Theresa, 1818, 3 86 1 PIEDMONT. Pistole, coined since 1785, half in proportion, 6 41 1 Sequin, half in proportion, 2 28 Carlino, coined since 1785; half in proportion, 27 34 Piece of 20 francs, called Ma- seneo, 3 56 4 POLAND. Ducat, 2 27 6 PORTUGAL. Dobraon, 32 70 6 Dobra, 17 30 1 Johannes, 17 06 4 Moi'lore, half in pioportion, 6 65 7 Piece of 16 Testooos, or 1600 Rees, 2 12 1 Old Crusado, of 400 Rees, 68 5 New do., 480 do., 63 5 Milree. coined in 1776, 78 PRUSSIA. Ducat, 1748, 8 27 9 do., 1787, 8 86 7 see DOODBOniAUIS. tM3 Namu op Ooma 4* en. Frederick, double, 1769, 7 95 5 do. do., 1800, 7 95 1 do. ■liigle, 1773, 3 99 7 do. do., 1800,. 8 07 5 ROME. Sequin, coined aiiioe mO^ 2 25 I SoudoofBepubUc, 16 811 RUSSIA. Ducat, 1796, 2 29 do., 1763, 2.26 Gold Ruble, 1756, 96 do., 1799. 78 do. Pollin, 1777, 35 Imperial, 1801, 7 82 Half do., 1801, 3 93 fe JIDINIA. Carl i no, lialf m proportion, SnXONY. Ducat, 1784, do., 1797, Augustus, n'54, do., 1784, SICILY. Ounce, 1751, 2 50 4 Double do., 1768, 6 04 4 SPAIN.' Doubloon. 1772, double and sin- gle, ana shares in proportion, 16 02 S Doubloon, 16 53 5 Pistole, 3 88 4 9 47 2 226 7 2 27 9 3-^2 5 3 97 4 N OF Coma. d. cm. -!■*" Coronilfa, Gold Dollar, or Vin> tern, 1801, 98 3 SWEDEN. Ducat, 2 SJ* 6 SWITZERLAND. Pistole of Hehmit BepoUte, 1800. 66 TREVEft Ducat. B2B TURKEY. Sequin FonducU, of Conftantl* nople, 1773, 1 tt do., 1789, 1 84 Half Miaseir, 1818, 52 Sequin- Fonducli, I 83 ^ Tesimieebielcblelc, 3 02 8 TUSCANY. Zechino, or Sequin, 2 31 8 Ruspone of the kingdom of Btruria, 8 93 8 VENICE. Zechino, or Sequin, duurw in prop ortion, 8 tt • WIRTEMBUBO; Carolin, 4 89 8 Ducat, 2 28 I ZURICH. Ducat, double and half In pn^ portion, S 91 7 ^tmttm 4«^ DUODECIMALS. V 9©S. Duodecimals are fractions of a foot. The word is derived from the Latin word duodecim, which signified twelve. A foot, instead of being divided decimally into ten equal parts, is divided duodedvudly into twelve equal parts, cadled primes, marked thus ('). Again, each of these parts 's conceived to be divided into twelve other equal parts, called seconds, ("). In like manner, each second is conceived to be divided into twelve equal parts, called thirds ( "' ) ; each third into twelve equal parts, called fourths ("") ; and so on to any extent. In this way of dividing a foot; it is obvious, that V r 1" 1" prime is *. ..... second is -^ of -j^j-, .... third is 1^ of 1^ of 1^, . . fourth is tV of tV of tV of -jV* fifth isTVof-iVof AofiVof A» -^ of a foot. yJ J of a foot. iVay of a foot Tu^TB of a foot. jTrsWy of a foot, &q T204 PUODECIMALS. 29f 12"" fa arths make 12"' thirds . . . 12" seconds . . 12' primes, . . 1"' third, 1" second, 1' prime, 1 foot >t. Note 1 . —The marks, ', ", '^, *", &c., which distinguish the different I arts, are called the indices of the parts or denominations. NoTs 2. — The divisions of a unit in duodecimals are uniform, just as in decimal fractions, with this difference : they decrease in a twelvt'/old propor- tion, 12 of a lower denominatioD making I of a higher. Operations in them are consetjuently the same as in whole numbers or decimids, except that 12 is the carrying number instead of 10. Multiplication of Duodecimals. IT 904* Duodecimals are used in measuring iurfaui and 1. How many square feet in a board 16 feet 7 inches long, and 1 foot 3 inches wide ? Note. — Length X breadth ca superficial contents, (IT 48.) OPERATION. ft. Length, 16 7' Breadth, 1 3' 4 16 r 7' 9 // Solution. — 7 inches, or primes, os ^ of a foot, and 3 inches as -^ of a foot ; conse- quently, the product of 7' X 3' =: i^ of a foot, that is, 2V' =i V ajid 9" j wherefore we set down the 9", and reserve the V to be car ried forward to its proper place. To multi- ply 16 feet by 3' is to take A of -^ = tf» that is, 48' J and the 1' whibh we reserved makes 49', ^ 4 feet 1' j we therefore set down the 1', and carry forward the 4 feet to its proper place. Then, raultipl3ring the multi- plicand by the 1 foot in the multiplier, and adding the two products to- gether, we obtain the Afiswer, 20 feet 8' and 9^. Note 1 . — In all cases the prodwct of any two denominations wiM always be of the denomination denoted by the sum of their ikdices. Thus, in the above Ans. 20 8' 9" Questions. — IT 203* What are duodecimals 7 Elxplain the duodecimal divisions and subdivisions of a foot. Repeat the table. What are indices? What part of a foot is l"? — 1"? — l'*? — 1*^7 — I'**? What difference between the decimal and duodecimal divisions of a unit 7 How are operation! flii duodecimals performed 7 23d DUODECniALS. 7204 2. How many solid feet Ji a block 15 ft. 8' long, 1 ft. 5' wide, and 1 ft. 4' thick ? OPERATION. Length, 15 8' Breadth, 1 5' The length multiplied by the breailth, and that product by the thickness, gives the aoUd conUnta, (IT 61.) 6 6' 4" 15 8' '». 2^ 2' 4" 1 4' 7 4' 9" 4'" 22 2' 4" /'/ Am. 29 7' 1" 4 Hence, To multiply diwdecimals, RULE* I. Write the multiplier under the multiplicand, like de- nominations under like, and in multiplying, remember that the product of any two denominations will be of that denomi- nation denoted by* the su7n of their indices. II. Add the several products together, and their sum will be the product required. EXAMPLES FOR PRACTICE* 3. How many square feet in a stock of 15 boards, each of which is /2 ft. 8' in length, and 13' wide? Ans. 205 ft. K^. 4. What is the product of 371 ft. 2' 6" multiplied by 181 ft. 1' O''? Ans. 67242 ft. IC 1" 4'" 6"^. 5. There is a room plastered, the compass of which is 47 ft. 3', and the bight 7 ft 6' ; what are the contents 1 Ans, 39 yds. 3 ft. 4' 6". 6. What will it cost to pave a court yard, 26 ft. 8' long by 24 ft. 9 wide, at $*90 per square yard? Ans. $66. 7. There is a house containing two rooms, each 16 ft. by 15 ft. 4'; a hall 24 ft. by laft. 6'; three bed-rooms, each 11 ft. 4' by 8 ft. ; a pantry 7 ft. by 9 ft. 6' ; a kitchen 14 ft. 2' by 18 ft., and two cham- Qnestions. — IT 20 !• For what are duodecimals used ? Of what de- nomination is the product of any two denominations ? Repeat the rule for the multiplication of duodecimals. How do you carry frrm one denomination to anoiher? How is masons' work estimated? What is understood by girt, ami for wJiat used? • •■ I ♦ • T 205. DUODECIMAL& 259 bers, each 16 ft. by 2C \.Qf; what did the work of iloorin|: cost, at $'03 per square foot? Ans. $39'95. Note 2. —Masons' wxk is esthnated by the perch of 16( feet in length, 1^ feet in width, and 1 foot in hight. A perch contains 24 '76 cubic feet if any wail be U feet thiclc, its contents in perches may be found by dividing its su- perficial contents by 16| : but if it be any other thickness than U feet, its cubic contents must be diYidea by 24<75, (=24$,) to reduce it to perches. Joiners, painters, plasterers, brick-layers, and masons, make no allowance for windows, doors, &c. Brick-layers and masons make no allowance for comers to the walls of houses, cellars, &c., but estimate their work by the girt, that is, the length of the wall on the outside. 8. The side walls of a cellar are each 32 ft. 6' long, the end walls 24 ft. 6^ and the whole are 7 ft. high, and 1|| ft. thick ; how many perches of stone are required, allowing- nothing for waste, and for how many must the mason be paid ? A S ^^^ perches in the wall. ( The mason must be paid for 48^ perches. 9. How many cord feet of wood m a load 7 feet long, 3 feet wide, and 3 ^t 4 inches high, and what wiU it cost at $'40 per cord foot? Ans. 4i cord feet, and it will cost $1*75. 10. How much wood in a load 10 ft. in length, 3 ft. 9^ in width, and 4 ft. S' in hight? and what will it cost at $1*92 per cordi Ans. 1 cord and 2f^ cord feet, and it will cost $2'62|. IT SOS* By some surveyors of wood, dimensions are taken in feet and decimals of a foot. For this purpose, make a rule or scale 4 feet long, and divide it into feet, and each foot into ten equal parts. Such a rule will be found very convenient for surveyors of wood and of lumber, for painters, joiners, &c. ; for the dimensions taken by it being in feet and decimals of a foot, the casts will be no other than so many operations in decimal fractions. 1. How many square feet in a hearth stone, which, by a rule, as above described, measures 4*5 feet in length, and 2*6 feet in width ^ and what will be its cost, at 75 cents per square foot? Ans, 11*7 feet ; and it will cost $8*775. 2. How many cords in a load of wood, 7*5 feet in length, 3*6 feet m width, and 4*8 in hight? Ans, I cord 1^ cu. ft. 3. How many cord feet in a load of wood 10 feet long, 3*4 feet wide, and 3*5 high? Ans, 7-/^. Qnestions* — ^ 205* How do some surveyors of wood take dimen- sions? Explain tha rule used in measuring. How are dimensions taken by i; estimated? L_. mVCLUTKHI. INVOLUTION. Ftrapoaer. T 9WI. Three feet in length (T 111) are a yard, lioear measure ; 3 in lengtb and 3 la width, 3x3=9 square feet, are a yard, Seamd power, square measure j 3 in length, 3 in width, and □ 3 in hight, 3x3 X 3 = 27 aoUd feet, aie a yard, cubic measure, (11113.) When a number, as 3, is multiplied into itself, and the product by the original number, and so on, the series of numbers produced are tailed powers, and llie process of producing them is called InoeitUion. Third voaer '^'** ^"^ number, represented by a line, is ■^^^' called the first power, or root ; the second, tq>- resented by a square, is called the square, or 3d power ; the third, rej)resentBd by a cube, is called the cube, or .3d power. The 4th power of 3 is 3 times the 3d power, 3 blocks like that employsd to r^resent the 3d power, and may be rep- resented by a figure 3 times as large, that is, 3 feet wide, 3 feet high, and 9 feet long. The 5th power, by 3 times such a figure, ta one 3 higb,0 wide, and 9 long. The 6th power, 3 times this, by a figure 9 long, 9 wide, and 9 high, or a cube. <;.wi «™™ Thus it may be shown that die ^"^"f"^- 9th, 12th, Ifta, 18th, tc., powers, may be represented by cubes ; the 7th, 10th, 13th, 16th, &c., by %- ures having greater length than width and hight; the 8th, Uth, 14th, 17th, &c., by figures having greater length ' and width than higM- To involve a number, take it aa B factor as many times as is indicated by the required power. Nan. 1 — Tha aambet denotii^ tha power i* oiled the indet, <• s^p*- mmti thai, t* dwiotei ibai e It niaed or uiTolTed to the <ih pover. tim, nnroL JVKNf . mt IQCAMPI^BS FOR PRACnCB. 1. What is the squaife, or 2i pomm^ •£ 7! S. What is the square of 30 ? 3. WbaiistheaquaiaofiOOOt 4. What is the cube, or 3d power« of 4 1 5. What u the cube of 800? 6. What is the 4th power of 00! 7. What is the square of 1 ? a Whatisthecvbeofl! 0. What is the squsxe of } ! 10. Whst is the eiOw of }t 11. What is the square «f if 13. What is the square of V5 ? — 13. What is the 6(h power of 1<3 ? 14. InvolYd 2^ to the 4th power. Ans, 40. Ans, 900. Ans. ia0000#0. Ans. 64. Ans. 512000000. Ans, 12960000. of3t of4? Ans. 1, 4, 9, and 16. of 3! of 41 Ans. 1, 8, 27, and 64. -of^l of J1 Ans, I, a, and |}. af^l ofjl ^^' ]fr» ih^ wd Hi- the 5th power of 1 1 4^5. ^ and -g^. — the cube ? iiii5. 2<S5, and 3<375. Ans. 9*985984. NoTB 2. — A mixed number, like the abore, may be reduced to sn im- proper fraction before involving ; thus, 2i s= |- ; or it may be reduced to a dedmal; thus, 2i ssa 2*25. 15. What is the square of 4^ ? Ans. ^4|a = 23|| . 16. Whs^ is the value of 7S that is, the 4th power of 7? Ans. 2401. 17. How much is 9^1 6M 10<? Ans. 729, 7776, 10000. 18. How much is 2M 3«! 4«! 5^ 6*1 10*1 Ans. to the hist, 100000000. Note 3.— The povers of the sine digits, from the first power to the filth may be seen in the fo^owing TABId& Roots . . or let Powers 1 2) 3 4 6 6t ^1 8 9 9r(uares or 2d Powers I 4| 9 16 25 36 49 64 t 81 Cubes . . j or 3d Powers nn d| iif 64 12^ 216 343 512 1 729 Biquadrates | or 4th Powers 1 1^1 ^1 256 6ii5 1296 2401 40961 6561 Sursoiids . | or 5lh Powers { 1 32 1 243 1024 ) 3125 | 7776 | 16807 32768 1 59049 Qnestions. — IT 306« What are powers? How is the first power rep- resented? Why is the second power called the square? Why the third called the cube 7 How is the fourth power represented ? —the fifth 7 — the siith? — thirtenth? — the fonrteenth? — the twenty-first? — the twenty- third ? — the twenty-fifth ? What is involution? Hrw is a number involved to any power? What is the index, and how written ? How is a mixed nam- ber iaf<idved1 S6B EVOLUTION. IT 207, 20a EVOLUTION. VfWfm Evolution, or the extracting of roots, is the method of finding the root of any power or number. The root, as we have seen, is that number, which, by a continual multiplication into itself, produces the given power, and to find the square root of a number (one side of a square when the contents are given) is to find a number, which, being squared, will produce the given number ; to find the cube root of-a number (the length of one side of a cubic body when the solid contents are given) is to find a number, which, being cubed or involved to the 3d power, will produce the given number : thus, the square root of 144 is 12, because 12^ =: 144 ; and the cube root of 343 is 7, because 7', that is, 7 X 7 X 7 = 343 ; and so of other numbers. NoTB. — Although there is no Bumber which will not produce a perfect power by involution, yet there are many numbers of which precise roots can never be obtained. But, by the help of dedmalSf we can approximate, or ap- proach, towards the root to any assiened decree of exactness. Numbers, whose precise roots cannot be obtainea, are called surd numbers, and thoso whose roots can be exactly obtained are called rational numbers. The square root is indicated by this character ^ placed before the num- ber ; the other roots by the same character, with the index of the root placed over it^ Thus, the square root of 16 is expressed >^16 ; and the cube root of 27 is expressed ^ 27 ; and the 6th root of 7776, ^7767. When the power is expressed by several numbers, with the sign + or — between them, a line, or virtculum^ is drawn firom th e top o f the sign over all the parts of it ; thus, the square root of 21 —5 is >^21— 5. Extraction of the Square Root. 9 IT 908* 1. Supposing a man has 625 yards of carpet- ing, a yard wide, what is the length of one side of a square room, the floor of which the carpeting will cover? that is, what is one side of a square, which contains 625 square yards? Solution. — We may find one side of a square containing 625 square yards, that is, the square root of 625, by a sort of trial ; and, '■ — — Qnestions* — IT 20T. What is evolution? What is a root? — tne square root, and how found? — the cube root, and how found? GiTe exam- Ijles. What do you say of perfect powers and perfect roots ? Give the dis- tinction between surd and rational numbers. How is the square root indi* cated ? — the cube root ? Describe the manner of using the Tinculum. r 5208. EVOLUTION. S63 OFERATION 625(2 4 225 Ist. We wil endeayor to ascenain how many ilgares tbeic wiH be in the root. This -we can easily do, by pointiog off the number, from units, into periods of two figures each ; for the square of and root always con- tains just twice as many, or one ligure less than twice as many figures, as are in the root. The square of 3 (3 X 3 »* 9) contains 1 figare , the square of 4 (4 X 4» 16) contains 2 figures ; the square of 9 (9 X 9 a 1) contains 2 figures ; the square of 10 (10 X 10 = 100) contains 3 figures ; the square of 32 (32 X 32 s= 1024) contains 4 figures j the square of 99 (99 X 99 =» 9801) contains 4 figures; the square of 100 (100 X 100 = 10000) contains 5 fig- ares, and so of any number. Pointing off the numlxir, we find *that the root will consist of two figures, a ten and a unit. 2d. We will now seek for the first fig- ure, that is, for the tens of the root, which we must extract from the left hand peri- od, 6, (hundreds.) The greatest square in 6 (hundreds) we find to be 4, (hun- dreds,) the root of which is 2, (tens, =» 20 ; therefore, we set 2 Aens; in the root. Smce the root is one sim of a square, let us form a square, (A, Fig. I.,) each side of which shall be regarded 2 tens, s= 20 yards long. The contents of this square are 20 X 20 = 400 yards, now disposed of, and which, consequently, are to be deducted from the whole .number of yards, (625,) leaving 225 yards. This de- duction is most readily performed by subtracting the square number, 4, (hundreds,) or the square of 2, (tens,) from the period 6, (hundreds,) and bringing down the next period to the remainder, making 225. 3d. The square A is now to be enlarged by the addition of ih6 225 remaining yards ; and in order that the figure may retain its sq^iare form^ the addition must be made on two sides. Now, if the 225 yards be divided by the length of the troo sides, (20-|-20 = 40,) the quotient will be the breadth of this new addition of 225 yards to the sides c d and i c of the square A. But our root already found, a= 2 tens, is the length of one side of the figure A J we therefore take double this root, = 4 tens, for a divisor. The divisor, 4, (tens,) is in reality 40, and we are to seek how many times 40 is contained in 225, or, which is the same thing, we may seek how many times 4 ^tens) is contained in 22, (tens,) rejecting the right hand figure of the dividend, because we have rejected the cipher in the Qnestions* — IT 308* How may one side of a sc^uare, when the contents are given, be found ? Why, in the trial, must we point off the number into periods of two figures each ? Illustrate. Why is the first root fisfure 2 tens ? Of what number is 2 tens the root? What may now be formed? How large? How must it be increased ? What will be the divisor? — tlie dividend? — the quotient? Why is the divisor too small? What is the entire divisor? How are the contents of the addition found? What does Pig. II. represent? What is the first method of proof ? — the second method? Fig. L d ^ A 20 80 20 400 2(r ., 904 MVWOWXff. ViN9. ^mMMtlQM^iSBVnWV M>. • • 625(25 4 45)225 225 Fig. n. 80 yds. S yds. i i 20 6 B 5 100 D J 26 d A 20 20 400 C c 20 6 100 a b at 1 at "I- 20 yds. 6 yds. The proof may be «een by adding figure, thus : — The square A contains 400 yards. The figure B « 100 « « « C " 100 D " 25 u divisor. Wo find our quotient, tiuU iSf the breadih of the addi- tion to be 5 yajtls ; but, if we look at Fig. li., we shall perceive that this addition of 5 yards to the two sides does not complete the square : for there is still wanting, in the corner J), a small square, each side of which is equal to this last quotient, 5 ; we . must, therefore, add this quotient, 5, to the divisor, 40, that is, place it at the right hcmd of the 4, (tens,) making the length of the whole addition formed by 225 square yards, 45 yards ; and then the whole length of the addition, 45 yeurds, multiplied by 5, the number of yards in the width, will give the contents of the whole addition around the sides of the figure A, which, in this case, be* ing 225 yards, the same as our dividend, we have no remainder, and the work is done. Conse- quently, Fig. II. represents the floor of a square room, 25 yards on a side, which 625 square yards of carpeting will exactly cover, together the several parts of the Or we may prove it by mvolu- tion, thu8:»25X25s=625, before. It a it Proof, 625 « IT 909. From this example and tUtutration toe derive the foUowing general RUUB FOB THE EXTBACTION OF THE SQUARE BOOT. I. Point off the given number into periods of two figures each, by putting a dot over the units, another over the hun- dreds, and so on. These dots show the number of figures of which the root will consist. II. Find the greatest square number in the left hand pe- riod, and write its root as a quotient in division. Subtract the square number from the left hand period, and to the remain- der bring down the next period for a dividend. S209. EVOLUTION. 265 m. Double the root already found for a divisor ; seeK how many times the divisor is contained in the dividend, excepting the right hand figure, and place the result m the root, and also at the right hand of the divisor ; the divisor thus in- creased will be the length of the whole addition now made to two sides of the square ; multiply the divisor, or length of the addition, by the last figure of the root, (the breadth of the addition,) and subtract the contents of Uie addition thus ob- tained from the dividend, and to the remainder bring down the next period for a new dividend. IV. Double the root already found for a new divisor, and continue the operation as before, until all the periods are brought down. Note 1. — As the value of figures, whether integers or decimais, b deter- mined by their distance from the place of units, we must always begin at units* place to point off the given number, and if it be a mixed number, we mast poiat it off both ways from units, and if there be but one figure in any period of decimals, a cipher must be added to it. And as the root must always- consist of as many integers and decimals as there are periods belong- ing to each in the given number, when it is necessary to carry the operation to a greater degree of exactness by decimals in the root, ai\er all the periods are brought down, two ciphers, a whole period, must be annexed for every decimal figure which we would obtain in the root. SXAMPIiKS FOR PRACTICE* 3. What is the sqaare root of 61504 ? ILLUSTRATION. 240 X B = 1920 64 200X40=* 8000 40 40 1600 o X 00 II i • 200 1 200 ^^ o o X o II 00 o o o 40000 1 200-1-404-8 OPERATION. 61504(248 Am 4 44)215 176 488)3904 3904 0000 The pupil will easily illustrate the operation by the annexed diagram. Proof. — 40000 -f 8000 + 8000 + 1600 + 1920 + 1920 + 64 « 61504, or 248 X 248 = 61504. Qnestions. — IT a09« What is the seneral rule for extracting the square root? Where must we begin to point on? What is done when one decimal place is wanting? How is the operation continued, when all the periods are Drought down? Why cannot the precise root be lound when there is a re- mainder? How is the root of a fraction obtained? Why? What is done when Uie terms of the fracticm are not exact squares ? 23 M6 EVOLUTION. IT 210. 3. What is the square root of 43264 ? Ans, 208. 4. What is the square root of 998001 1 Ans. 999. 5. What is the square root of 234'09 ! Ans. 15'3^ 6. What is the square root of 964'5192360241 ! Ans. 31*05671. ^ 7. What is the square root of '001296? Ans. *036. 8. What is the square root of *2916 ? Ans. *64. 9. What is the square root of 36372961 1 Ans. 6031. 10. What is the square root of 164 1 Ans. 12'8-f-. Note 2. — In the last example, there was a remainder, after a^ the fibres were brought down. In such cases, the {)recise root can never be obtained. For, as the operation is continued by annexing ciphers, the last fieure of every dividend must be a cipher. But the root figure obtained from this dividend, is also placed at the right hand of the divisor, and consequently is multiplied into itself, and the last>.fifure of the product placed under the ci]^er, which is the last figure of the dividend, to be subtracted from it. And as the product of uo one of the significant figures ends in a cipher, there will always be a re- mainder. 1 1 . What is the square root of 3 ? Ans. 1*73 4-. 12. What is the square root of 10? Ans. 3*16-4-. 13. What is the square root of 184*2 ? Ans. 13*57-4-. 14. What is the square root of ^? Note 3. — Since, from the rule for multiplying one fraction by another, a fraction is involved by involving its numerator and its denominator, the root cf a fraction is obtained by finding the root of its numerator, and of its denom- '^*'*''- Ans. |. 15. What is the square root of ^ ? Afis. ^. 16. What is the square root of -^^ ? Ans. ^. 17. What is the square root of t^? Ans. ^= J. 18. What is the square root of 204 ? Ans. 4^. Note 4. — When the numerator and denominator are not exact sqiuxreSf tht fraction may be reduced to a decimal, and the approximcUe root found. 19. What is the square root of i = *75 ? j5is. *866 +. 20. What is the square root of J^. Ans. *912 -j-. PRACTICAIi £X£RCIS£S IN THB EXTRACTION OF THE SQUARE ROOT. V 310« 1. A general has 4096 men ; how many must he place m rank and file to form them into a square? Ans. 64. 2. If a square field contains 2025 square rods, how many rods does it measure on each side? Ans. 45 rods. 3. How many trees in each row of a square orchard containing 6625 trees? Ans. 75. 4. There is a circle whose areay or superficial contents, is 5184 feet ; what will be the length of the side of a square of equal area ? >x/^184 = 72 feet, Ans. 5. A has two fields, one containing 40 acres, and the other con- taining 50 acres, for which B ofifers hmi a square field containing the r T 210. KVOLunoN. 267 stme number of acres as both of these ; how many rods must eaeh side of this field measare? Ans, 120 rods. 6. If a certain square field measure 20 rods on each side, how much will the side of a square field measure, o ntaining 4 times as n^nch^ V20 X 20 X 4 = 40 rods, Ans. 7. If the side of a square be 5 feet, what will be the side of one 4 times as large? 9 times as large? 16 times as large! ■ 25 times as large? 36 times as large? Answers, 10 ft, ; 15 ft. ; 20 ft. ; 25 ft., and 36 ft. 8. It is required to lay out 288 rods of land in the form of a paral- lelogram, which shall be twice as many rods in length as it is in widih. Note 1. — If the field be dinded in the middle, it will fonn two equal ■quares. .^715. 24 rods long, and 12 rods wide. 9. I would set out, at equal distances, 784 apple trees, so that my orchaiQ may be 4 times as long as it is broad ; how many rows of trees must 1 have, and how many trees in each row ? "^Ans, 14 rows, and 56 trees in eaeh row. 10. There is an oblong piece of land, containing 192 square rods, of which the width is { as much as the length ; required its dimen- aorf&. ^ Ans, 16 by 12. 11. There is a circle, whose diameter is 4 inches; what is the diameter of a circle 9 times as large ? NoTs 2. — A square 4 inches on one side, contains 16 s(juare Inches ; one tvfice as long, or 8 inches on each side, contains 64 square inches, 4 times 16 ; one 3 times as long, or 12 inches on each side, contains 144 =9 times 16 square inches. It may also be shown by geometry, that if the diameter of a circle be doubled, its contents will be increased 4 times j if the diameter be trebled, the contents viiW be increased 9 times. That is, the contents of squares are in proportion to the squares of their sides, and the contents of cir- cles are in proportion to the squares of their diameters. Hence, to perform the above example, square the diameter, multiply the square by 9, and extract the square root of the product. Ans. 12 inches. 12. There are two circular ponds in a gentleman's pleasure ground ; the diameter of the less is 100 feet, and the greater is 3 times as large ; what is its diameter ? Ans. 173*2 -f- feet. 13. If the diameter of a circle be 12 inches, what is the diameter of one i as large? Ans. 6 inches. 14. A carpenter has a large wooden square; one part of it is 4 feet long, and the other part 3 feet long ; what is the length of a pole« which will just reach from one end to the other? Note 3. — A figure of 3 sides is called a triangle, and if Pig. 1. one of the comers be a square comer , or right angle, like 4. the angle at B in the annexed figup, it is called a right angled triangle. It is proved by a geometrical demonstra- tion that the square contents of a square formed on the longest side, A C, are equal to the square contents of the two squares, one formed on each of the other two sides, A B, ana C B. Thus, Vi^. 2, a square formed on A B, the shortest side, will co' tain 9 square feet, the square on C B EVOLimON. tsio. «1U oontaln IS Bqinn feet, e + lS^U tquare feet, in bctb equsres. The uuara DQ A C contains 35 enM wuarei o! dm ■ams size na the w^uares on Ibe olhei twu ■ide9 Hie divided lata, or 35 square feet, ■od the squaie root of 25 will be the lenglll of the loDgeat aide, or, An*., 6 feet. HoDce, if the length of the (»D short sides HTC ^ven, ryuare each, add the tquarea together, and extract the suuare tfttare root q/ the earn f the root wili Be the ieiL^/i oftAe loBff tide. If tiie long side, and one of the short sides are given, eipun-e each, lubtratt the rire of the thort side from the iquareiif long side; the agaare root of the tv- Tnotnder vt3 be the oiher short iide. 15. If, from the eonier of a equare room, 6 feet be messufed off one way, tad 8 feet the other way, along' the sidea of ^e locmt, what will be the length of a pole teaching from point to point ! Am. 10 feet. 16. A wall is 33 feet high, and a ditch before it ia 24 feet wide ; what is the length of a ladder that will reach from^e top of the wall U> the opposite side of the dilohl Xtu. 40 feet, 17. If the ladder be 40 feet, and the wall 32 feet, what is the width of the ditch ? Aia. 24 feet. 18. The ladder and ditch given, required the wall. Ans. 33 fbet. la. The distance between the lower ends of two equal rafters is 33 feet, and the hight of the ridge, above the beam on which they stand, is 13 feet ; required the length of each rafter. Ans. SO feet. 30. There is a building 30 feet in length and 23 feet ui width, and the eaves project beyond the wall 1 foot on every side ; the roof ter- minatca in a point at the centre of the building, and is there supported by a post, the top of which is ID feet above the beams on which the rafters rest ; what is the distance from the foot of the post to the cor- nerfl of the eaves? and what is the length of a railer, reaching to the middle of one sidef a rafter reaching to the middle of one ettdf and a rafter reaching to the eorneri of the eaves % .Anjuwri, in order, soft. ; I5'62 + ft.; l8'86+ft.; and29'36 + feet. 31. There is a £eld 800 rods long and 600 rods wide ; what is the distance between two opposite comers ? Ans. 1000 rods. 23. There is a square field ccmtaining 00 actee ; hew many rods itenu 01 a square to il aOect the conlenn or circJea to double or treble their diamelera 7 How will you find the dianv rifriii angled triangle 7 What is said of (bo squares on its aides 'J How shown DvPig. 21 When bolt short sides are given, how do jou find the longslde? Wh«i the Id^ aide, and one short side are given, hov do you find the ctherl ^211. EVOLUTION, 269 ID length is each ride of the Geld! and how many rods apart an tliB opposite comers \ Answers, 120 rods, and 1B9'74- rods. 23. There is a square field containiiig 10 acres ; what distance ii the cenue from each comeil Arts. 28'S84-rods. Extraction of the Cobe Root. T 311< 1- How many feet in length is each side of a cubic block, containing 125 solid feet! SoLQTioK. — As the solid conlents of a cubical body are fbtmd, when one side is known, by inyolving the side lo the Ihiri power, or cnbe, (IT 206,) so when ike solid conlenw are known, we flm the length of one side by eiiracting ihe cube root, a number, which, taken be h fac- " ■ s, wiil prodace the given number, f 1[ 207,} The cube root of J, 1[206, lobeS. ,Ani. Sfeet. 125 we find by Inspeclion, or by the table, 1 4(2 5824 Fig. I. S. What is the side of a cubic Uoek, coataininff 64 solid feet? — 37 solid teetl 216 solid feet? 513 8ohdfeett Aruteeri, 4 ft., 3 ft., 6 ft., and 8 ft. 3. Supposing a man has 13824 feet of timber, in separate blocks of 1 cubic foot each ; he wishes to pile them up in a cubic pile ; what will be the length of each side of such a OPERATION, Solution. — It is evident ihal, as in the former examples, wc must find the length of one side of a cubical pile which 13824 sach blocks will make by eilracling the cube root of 1382J. But this number is so large, that we cannot so easily find Ihe rcot as in Ihe former examples ; — we wlI C 20 _ D endeavor, however, to do it by a »Mt Isi. We will try to ascertain the number of figures, of which the root will consist. This wa may do by pointing the nomber off into periods of tJiret figures each. For (he cube of any figure will contain 3 times as many, or 1 or 2 less than 3 times as many figures as the number its^f. The cube of 2 craitains 1 figure ; iBe cube of 5 contains 2 figures ; the cube 20 »f 9 comains 3 figures j the cube of 10 ^ — contains 4 figures, and so on. ^'^ Pointing off, we see that the root ^ will consist of two figures, a ten and 8000 /ue, axiteflM. aimil. Let Qs, then, seek for (he first 23* m EvomnoN. HSU. flgan, or (ens nf the root, which most be eiCiBcted from the left hand period, 13, (lhoii£B.iids.] The grealeft cube in 13 (thousands) we Gtid ig intpeaion, or bf the lable of poteen, to l>e 6, (thousands,) the root of which is a, {tens ;} therefore, we place 2 (lens) in the root. As the root is one side of a cube, let ua form a cube, (Fi^. I,,) each side of which shall be regarded 20 feet, expressed by Ibe root now ubtsined. The -.ontents of this cube are 20 X -1 X ^C = 8000 solid feet, which are now disposed of, and which, consequently, are to be deducted from the whole ' number of feet, 138S4. 8000 taken from 13824 leave 3S21 feet. This deduction is most readily performed by subtracting the cubic number, 8, ca' the cube of 2, (the figure of the root already found,) from the perii>d 13, (tliousands,) and britiging down the neit period by the side of the remainder, matting 5824, as before. 3d. The cubic pile A D is now to be enlarged by the addition of 5624 'Solid feel, and, in onlei lo preserve the cubic form of the pile, the addi- tion must be made on one half of its sides, that is, on 3 sides, a, h, and e. Now as each side is 20 feet square, its square contents aie 400 Snare feel, and the square contents of the 3 sides are 1301) square feet, ence, an addition of I foot thich would require 1200 solid feet, and dividing 5834 solid feet by 1200 OPEBATION - CONTINUED. 13824(24 5oo(. Divisor, 1300)5824 Dividend. solid feel, the contents of the ad- ditioti 1 foot thick, and we get the tbicltness of I he addition. It will be seen that the quotient figure must not always be as large as it can be. There might be enough, for InslaDce, to make the three Sjddi- tioiis now under consideration 5 feet thick, when there would not then be enough remaining lo complete the additions. The divisor, 1200, is contained in the dividend 4 times ; conse- quently, 4 feet is the thickness of the addition made to each of the three sides, a, b, c, and 4 X ^^09 = 4800, is the solid feet contained in these additions ; but there are still 1024 feet left, and if we look at Fig. II., we shall perceive that this addition to the 3 sides does not complete the cube; for there are deficiencies in ihe 3 comers, n, R, n. Now the Itagih of each of these defiritMki is the same as the length of taeh side, that is, 2 (1ens)=S0,.and their truKA and thickness are each equal to the !/al gvuiitnt figure, (4 ;) their con tents, therefore, or the number o) feet required lojUl these dencicn- cies, will be found by multiplying the sjuare of the last quotient fig- ure. W' J "16. by 20; 16 X 20— KVOLVTUm. 271 320 sotid fMt, reqiiirad for one de- ficiency, and mnliiplTing 320 bj 3, 320 X 3 —3^0 solid feel, Rqiuicd for the 3 deficiencies, n, n, n. Lot^D^ aL Fig. III., ve peiceive there is sull a deficiency in the cor- ner vhere the last blocki meet. Tbis deficiency is a cube, each aids of which is equal 1o the Inst quo- tient figure, 4. The cQbe of 4, Iherefore, (4X4X4™ <>*:) willbe the solid conlenis of this comer, which in Fig. tV. is seen filled. Now, the sunt of tbe$e several iddiiiOQS, iriz,, 4800 -|- 9ti0 + « ■=5634, will make the subtrahend, which, subtracted from the divi- dend, leaves no remainder, and the work is done. Fig. IV. shows the pile which 13824 solid blocks of one fool each would make, when laid together, and the root, 24, shows the length of one Bide of the pile. The cor- rectneas of the work m'/ he ascer- tained by cubing tue side now found, 24^ thus, 24 X ^4 X ^4 — 13324, the given nnmber ; or it may be proved by adding together the contents of all the several parts, 8000 » contents of Fig. T. 4800 =• addition to the sides, a, b, and t, Hg. I. 9G0 = addition to fill the deficiencies n, "> ", F>K- 11. 64 = addition to fill the comer, t, t, e, Ptg. 17. 13824 — contents of the whole pile, Fig. 17., 24 feet on each side. iple and iUttstratum toe IT 313. From the foregoing terive t/ie following Qnestions, — V3I1. How is the length of one aide of a cube found, Krhen the contenla are known"! Why, Ei. 3, is the nnmber pauilct off as it M? Hov many dfurea in the cubs af any number! lllustiate by culiine mmn numbers. What is 2, the dcst fisiire of the root? Of what is it th<^ rooti For what is the Eulitiaclion? What is to be done with the remain- derT On how many iid«s is il to be added, and why? What is the divisor, 13007 What is the object in dividing 7 The quotient exfinssea what? Why shoatd it not be made as lane as it can be 7 What additions ore next made, and what are the contents oreach 7 How are the comenis found 7 What de- ficiency yei lemnins, and how large 7 Of what parts of the lost figure does the iDblrahend conBiBt7 DeKcibe Fig, 1. 1 — Fig. II. i — Fig. III. ; - Fia. IV. How ii the work proved? 272 E5Y:>LUTiort. ir21S. RHUS FOR EXTRACTING THE CUBE ROOT. I. Place a point over the unit figure, and over every third figure at the left of the place of units, thereby separating the given number into as many periods as there will be figures in the root. II. Find the greatest complete cube number in the left hand period, and place its cube root in the quotient. III. Subtract the cube thus found from the period taken, and bring down to the remainder the next period for a divi dend. IV. Calling the quotient, or root figure now obtained, so many tens, multiply its square by 3, and use the product for a divisor. V. Seek how many times the divisor is contained in the dividend, and diminishing the quotient, if necessary, so that the whole subtrahend, when found, may not be greater than the dividend, place the result in the root ; then multiply the divisor b^ this root figure, and write the product under the dividend. VI. Multiply the square of this root figure by the former figure or figures of the root, regarded as so many tens, and the resulting product by 3, add the product thus obtained, together with the cube of the last quotient, to the former product for a subtrahend. VII. Subtract the subtrahend from the dividend, and to the remainder bring down the next period for a new dividend, with which proceed as before, till the work is finished. Note 1. — If it happens that the divisor is not contained in the dividend, a cipher must be put in the root, and the next period brought down for a divi- dend. Note 2. — The same rule must be observed for continuing the operatioii, and pointing off for decimals, as in extracting the square root. EXAMPIiBS FOR PRACTICE. 4. What is the cube root of 1860867 ? Questions — IT 21!^> What is the general rule? — note 17-* note 8? — note 3 ? 1 S13. BTOLUnOIf . MS OPERATION. 1860867(123 Am. 1 10* X 3 = 300)860>rf Dividend. 600 f^X 10X3 = 120 2" = 8 728 Jirst Subtrahend. 120* X 3 = 43200)132867 second Dividend. 129600 3»X 120x3= 3240 3»= 27 132867 second Subtrahend. 000000 6. What is the cube root of 373248 1 Ans, 73. 6. What is the cube root of 21024576 ? Ans. 276. 7. What is the cube root of 84*604519 ? Ans.. 4*39. 8. What is the cube root of *000343 ? Ans. *07. 9. What is the cube root of 2 ? Ans. 1*25+. 10. What is the cube root of ^ ? Ans. §. NoTB 3. — The cube root of a fraction is the cube root of the numeratoT diTided by the cube root of the denominator. (IT 20d.) 11. What is the cube root of ^^ ? Ans. |-. 12. What is the cube root of -^7^^ ? Ans. ^. 13. What is the cute root of -^xr? Ans. *125-|-. 14. What is the cube root of y^ ^ ^'*** i* PRACTICAIi EXERCISTES IN ESXTRACTING TH£ CUBB ROOT* % 313* 1. What is the side of a cubical mound, equal to one 288 feet long, 216 feet broad, and 48 feet high ? Ans. 144 feet. 2. There is a cubic box, one side of which is 2 feet ; how many solid feet does it contain ? Ans. 8 feet. 3. How many cubic feet in one 8 times as large? and what would be the length of one sid 3 ? Ans. 64 solid feet, and one side is 4 feet. S74 EVOLUTION. Y 214 4. There is a cubical box, one side of which is 5 feet , what would be ^e side of one containing 27 times as much 1 64 times as much! 125 times as much? Ans, 15, 20, and 25 feet. 5. There is a cubical box, measuring 1 foot on each side ; what it the side of a box 8 times as large? 27 times? 64 times? Ans. 2, 3, and 4 feet. Note. — It appears from the above examples that the sides of cubes are as t]ie cube roots of their solid contents, and their solid contents as the cubes oi their sides. It is also true that if a fflpbe or ball have a certain contents, the contents of one whose diameter is docble are 8 times as great, oi loiving a treble diameter are 27 times as ^^reat, and so on ; that is, the contents are pro- portional to the cubes of their diameters. The same proportion is true of the BinUlar aidest or of the diameters of all solid figures oi similar forms. 6. If a ball, weighing 4 pounds, be 3 inches in diameter, what wiL be the diameteit of a bidl of the same metal, weighing 32 pounds? 4 : 32 : : 3^ : 6^. Ans. 6 inches. 7. If a ball 6 inches in diameter weigh 32 pounds, what will be the weight of a ball 3 inches in diameter? Ans, 4 lbs. 8. If a globe of silver, 1 inch in diameter, be worth $6, what is the yalue of a globe 1 foot in diameter? Ans. $10368. 9- There are two globes ; one of them is 1 foot in diameter, and tlie other 40 feet in diameter ; how many of the smaller globes would it take to make 1 of the larger ? Ans, 64000. 10. If the diameter of the sun is 112 times as much as the diame- ter of the earth, how many globes like the earth would it take to make one as large as the sun ? Ans. 1404928. 11. If the planet Saturn is 1000 times as large as the earth, and the earth is 7900 miles in diameter, what is the diameter of Saturn? Ans. 79000 miles. 12. There are two planets of equal density ; the diameter of the less is to that of the larger as 2 to 9 ; what is the ratio of their solidi- ties? Ans, yf^ ; or, as 8 to 729. IT 314« Review of Involution and Evolution- Questions* — What is involution? What^re powers? How are the different powers represented ? How is a number involved ? What IS evolution ? What is a root ? How do you find the square rcot, oi the cube root of a number ? What is a rational, and what a surd num- ber ? How is the square root indicated ? — the cube root ? Give briefly the solution of the example in the extraction of the square root ; — rule. How are decimals pointed off? How is the o])eration continued, when there is a remainder? Why cannot the precise root be ascertained? How is the square root of a vulgar fraction found ? What is said of the relation between the sides and contents of squares ? — the diameters and Qvestions. — H 213* What proportion exisU between the sides of cubes, and their solid coutcnts ? Illustrate. What betweec the diameters of globes and their contents 7 If you increase the diameter of a hkH 6 timesi bow much are its contento increased? T315. ARrrHMBnCAL PROGRESSION. 296 contents of cirdes? — the squares on the sides of a rig^it-anc^ed trian* 0e? Repeat briefly the solution of the example in cabe root ; — the role. What is said of the relation of the sides of cubes to the ccmtents f — of the diameters of globes to their contents 7 KXBRCISES* 1 What is the difference of the contents of 6 fields, each 20 rods square, and 1 field 50 rods square? Arts, 100 square rods. 2. What is the difference between 56 cubical stacks of hay, each 10 feet on a side, and 1 stack 40 feet on a side t Ans, 8000 solid feet. 3. How many times larger is a circular pond, 1 mile in diameter, than one that is 40 rods in diameter? Ans. 64 times. 4. What is one side of a cubical pUe of wood which contains 4 eords? Arts. 8 feet. 5. What is one side of a cubical pile of bricks which will lay up the walls of a house 36 feet high and 16 inches thick for the first 13 feet, 13 inches the next 13, and 8 inches the upper 13, the house being 60 feet long and 34 wide on the outside, no allowances being made for windows, doors, &c. ! What are the solid contents ? .^115. to the last, 6613 cu. ft., 576 cu. in. NoTB. — The principal object in eTolution is to find one side of a square or of a cube, when the contents are known, or to extract the square and cube roots. There are method's of demonstrating these operations dififerent from those here given, which are preferable in some respects, but they are deficient in one important particular — intelligibleness to tnose for whom they are de- signed. In a " higher arithmetic" they mit^ht be appropriate. Other roots may be extracted arithmetically, but tne methods of demonstrat- ing the operations, even where any are given^ are difficult of comprehension. The fourth root, however, may be found by taking the square root of the square root, the sixth root by taking the square root of the cube root, and so of many other roots. Any root is easily taken by what are called logarithms, used in the more advanced departments of mathematics. ARITHMETICAL PROGRESSION. IT 91S. 1. A teamster starts with 6 barrels of flour ; he passes by 4 mills, at each of which he takes on 3 barrels ; now many barrels has he then ? SoLLTioN. — He has 8 barrels after the first addition, 11 after the second, 14 after the third, and 17 after the fourth. Ans. 17 barrels. 2. A peddler having 17 hats, sold 3 at each of 4 stores , how many had he left? Solution. — He had 14 after the first sale, 11 after the second. 8 after the third, and 5 after the fourth. Ans. 5 hats. A series of numbers increasing by a constant addition, or 276 ARITHMETICAL PROGRESSION. f 216. decreasing by a constant subtraction of the saitt* Aumber, 18 called an Arithmetical Progression, The first of the above examples is called an ascending, the second a descending series. Note 1 . — The numbers which form the series are called the terma of tHe series. The Jirat aud last terms are the extremes, and the other terms are called the means. There are five things in an arithmetical progression, any th^ee of which being given, the other two may be found : — 1st. Th&Jirst term. 2d. Thelosnerm. 3d. The number of terms. 4th. The common difference. 6th. The sum of all the terms. Note 2. — The common difference is the number added o: subtracted at one time. V 316« Om of the extremes^ the common difference^ and the number of terms being giveny to find the other extreme. 1. A man bought 100 yards of cloth, giving 4 cents for the first yard, 7 cents for the second, 10 cents for the third, and so on, with a common diflerence of 3 cents ; what was 'the cost of the last yard ? Solution. — We add 3 to 4 cents, (4 -|- 3 = 7,) lo get the price of the second yard, 3 to 7 to get the price of tne third yard, and so on, thus making 99 additions to 4, of 3 cents each ; or, we may take 3, 99 times, (the multiplication being a short way of performing the 99 additions,) and add the product to 4, for the price of the last yard, 3 X 99 j or, since either factor may be the multiplier, 99X3 = 297, and 44-297 = 301 cents, the price of the last yard. Arts. 301 cents. Note 1. — The prices, 4, 7, 10, 13 cents, &c., are an ascending series, which has as many terms as there are yards, namely, 100 ; 3 is the common difference, and 4 the first term, to which 99 times 3 niust he added to find the Erice of the last yard, or the last term. It is added 1 time less than the num- er of terms, since 4 is the price of the first yard without any addition. Hence, Tojbid the last term of an ascending series when the first term, common difference, and rvumher of terms are given, RULS. Multiply the common difference by the number of terms less one, to get the sum of the additions, and add this sum to the first term ; the amount will be the last term. Questions. — IT 215. How is Ex. 1 explained? — Ex. 2? What are the extremes ? — the means ? — the terms ? How many, and what things are there, of which, if three are given, the others may be found ? What is tn« common difference 7 IT 217. ARITHMETICAL PROGRESSION. 277 NoTS2. —If the same things are ^ven of a deacending aeres, we must cridently take the sum of the suhtractions from the first term to find the last. In the same manner we may find the first term of an ascending series when the last term and the other tilings named are given : but having tbes^ things l^ven of a descending series, we find the first term by the rule above for find- ing the last term of an ascending series. EXAMPIiKS FOR "PRACTICK. 2. There are 23 pieces of land, tl:e first containing 95 acres, the second 91, the third 87, and so on, decreasing by a common difiference of 4 ; what is the number of acres in the last piece 1 Ans. 7. 3. xTie first term of a series is 6, the common difiference is 3, and tlie number of terms is 57 ; what is the last term ? Ans. 174. 4. The last term of a series is 117, the common difiference is 8, and the number of terms is 15 ; what is the first term ? Ans. 5. 5. The last term is 6, the number of terms 21, and the common difiference 10 ; what is the first term 1 Ans. 206. Simple Interest by Progression. ^317* 1. A man puts out $10, at 6 per cent, simple interest ; to what does it amount in 20 years ? Solution. — The first sum is $10, the amount at the end of the first year is $10*60, at the end of the second year SI 1*20, increasing each year by the constant addition of S<60. Hence, simple interest is a case of arithmetical progression, the principal being the first term, the inter- est 'for one year being the common difference, the number of terms one more than the number of years, since there is one term, the principal, at the commencement of. the first year, and one term, the amount for a year, at its close, and the last term, which we wish to find, is the amount for the number of years. To find the last term, or this amount, multi- £ly the interest for 1 ijear by the number of years, (one less than the num- ;r of terms,) and add the product to the first term. Ans. $22. 2. Two lads, at 14 years of age, commence labor for themselves ; the one lays up nothing, but the other, by prudence, lays up $300 by the time he is 20 years old, which he puts out at 7 per cent, simple interest ; afterwards, each earns his living, and no more ; at the age of 70 the one is worth nothing, and comes upon public charity ; what is the other worth at that age? Ans, $1350. Qnestions* —IT 216« What things are given, and what are required, in ir21€.' How many eases may there he, and what are they? What are given ij. Ex. 1 7 How much is the first term increased to make the last? Why aio ouly 99 times 3 added ? Give the rule. To what case does it apply 7 What is done i i each case, when other things are given ? IT 217. H w does it appear that simple interest is a case of progression? What things, according to IT 216, are given, and what is required? Why is there one more term than the number of years ? Hon i^^mple interest pei^ formed by progression ? 24 ^78 ARITHMEITICAL PBOGRSSSI0N. f 218, 219. 3. What will a watch, purchased at 31 for $25, cost an indiTiflual by the time he is 75, reckoning nothing for repairs but simple interest at 6 per cent, on iho purchase money ? — — at 8 per cent. ? ^715. to the last, $133. IT 31 8* The extremes aTid the number of terms given, to find the common difference, 1. The prices of 100 yards are in arithmetical progression, the first being 4, the last being 301 cents ; what is the com- mon increase of price on each succeeding yard ? Solution. — As the first yard costs 4 cents, 297 cents have been added to 4 for the price of the last yard, at 99 times, and dividing the number added at 99 times by 99, wc get the number added at 1 tim^. Hence, RUIiB. Divide the whole number added or subtracted, by the num her of additions or subtractions, that is, the difference of the extremes by the number of terms less 1, and the quotient i? the number added or subtracted at one time, or the common difference. EXAMPLES FOR PRACTICB. 2. If the extremes be 5 and 605, and the number of terms 151, what is the common difference ? Ans, 4. 3. A man had 8 sons, whose ages differed alike ; the youngest wa? 10 years old, and the eldest 45 ; what was the' common difference of their ages? Ans. 5 years. Note. — If the extremes and common drfierence are given, we may find the number of terms by dividing the difference of the extremes by the common difierence, and adding 1 to the quotient. 4. The extremes are 5 and 1205, and the common difierence 8 ; what is the number of terms 1 Ans, 151. ^ 319« The extremes and the mimher of terms being given, to find the sum of aU the terms, 1. What is the amount of the ascending series, 3, 6, 7, 9 11, 13, 15, 17, 19? Solution. — The sum may be found by adding together the terms , but in an extended series this process would be tedious. We will there- fore seek for a shorter method ; and first, will write down the terms of C^aestions. — IT 218* What are given and what required, IT 218 ? Ex- plain how the common difi*er;nce is found, Ex. 1 . Give the rule. How is the number of terms found, wbsn Htm extremes and the conunon diffiBrenoe are given? 1 T2S0. AKITHltETICAL PROGRESSION. 279 the series in order, aB.d beginning with the last, write the terms of the same series under these, placing the last term under the first, the Dext to the last under the second, the third irom the last under the third, and 80 on, thus : — * 3 5 7 9 11 13 15 17 19 19 "17 15 13 11 9 7 5 3 2222 22 2222222222 22 Adding together each pair, we see that the sums are alike, and the amount of the whole is as many times 22, the first sum, as there are terms in either series, which is 9. 22 X 9 = 198, the number in both oeries, and 198 -^2 as 99 must be the sum of the first series, which we wish to find. But 22 is the sum of the extremes of the series j hence, token the extremes and the number of terms are given to find the sum oj the termsy RULE* Multiply the snm of the extremes by the number of terms, and half the product will be the sum of the terms. EXAMPLES FOR PRACTICE. 3. If the extremes be 5 and 605, and the number of terms 151, what is the sum of the series? Ans. 46055. 3. What is the sum of the first 100 numbers, in their natural order, that is, 1, 2, 3, 4, &c. 1 Ans, 5050. 4. How many times does a common clock strike in 12 hours ? Ans. 78. Annuities by Arithmetical Progression. IT 290. An annuity, (from the Latin word annus, mean- ing a year,) is a uniform sum, due at the end of every year. When payment is not made at the end of the year, the annu- ity is said to be in arrears, and the sums of the annuities should draw interest just as any other debts not paid when due. When on simple interest, the several years* annuities, with the interest on each, form an arithmetical progression, and the calculation to ascertain the whole sum due, is — finding the sum of an arithmetical series ; thus : — Qaestions. — IF 219. What How might the sim be found ? \ ihe process by wh ch a shorter method sfrom tnc additions, Ex. 1 ? What does the product express, and why is it divided by 2 ? What is the quotient 7 Why is 22 the sum of the extremes 1 OiTe the rule. 280 i^RTTHMETICAL PROGRESSION. ¥221. 1. A man, whose salary is $100 a year, does not receive anything till the end of 8 years ; what was then his due, sim- ple interest on the sums in arrears at 6 per cent. ? Solution. — The first year's salary not being paid till 7 years aft%i it is due, since it was due at the end i )f the first year, is on interest 7 years. The interest of $100, 7 years, is $42, and $100 + 42 = $142, which he should receive on account of his first year's salary. His sec- ond year's salary, on interest 6 years, will amount to $136, his third year's salary will amount to $130, and so on, decreasing uniformly by $6, the interest of $100 for a year, till the last year, when the salary, being paid at the end of the year for which it has accrued, will not be on interest, but will yield him $100. The sums, $142, $136, $130, $124, $118, $112, $106, $100, form a descending arithmetical progres- sion, and to find the sum due, multiply the sum of the extremes by the number of terms, and take half the product, thus : — $142 4- $100 = $242 ; and $242 X 8 = $1936, which -5- 2 = $968, Arts, KXAMPIiBS FOR PRACTICB. 2. A soldier of the revolution did not establish his claim to a pen- sion of $96 a year till 10 years after it should have begun ; what was then his due, simple interest on the sums in arrears at 6 per cent. ? Ans, $1219*20. 3. A man uses tobacco at an expense of $5 a year from the age oi 18 till the age of 79, when he dies, leaving to his heirs $300 ; what might he have left them if he had dispensed with the worse than use- less article, and loaned the money, which it cost him, at the end of each year, for 7 per cent., simple interest? Ans^ $1245*50. 4. A and B have the same income, but the expenses of A are $210 a year, and those of B are $250 ; at the end of 40 years B is worth $1500 ; what is A worth, having loaned what he saved more than B at 7 per cent, simple interest at the end of each year ? Ans, $5284. 5. Refer to T[ 164, Ex. 8 : what would the merchant gain if he continued in trade 31 years to borrow money instead of purchasing on credit, loaning the money saved at the end of each year at 7 per cent, simple interest? ^ws. $20 161 '70 nearly. EXKRCISBS. V !3Sm. 1. If a triangular piece of land, 30 rods in length, be 20 rods wide at one end, and come to a point at the other, what num- ber of square rods does it contain ? ^715. 300. Questions. — IT 220. What is an annuity ? Why so called ? When are annuiiies; in arrears ? Whj' should thev then draw interest? When will they form an arithmetical progression? What is the calculation to find the whole sum due? Why, Ex. I, was the first year's salary on interest 7 years? Show how long each year's salary W€is on interest. Why was not the last year's salary on interest ? Why do the sums form a descendirg series ? How nay the sum be found ? Why is not the number of terms ona more \han the number of years, as in V 217 ? T222. GEOMETRICAL PROGRESSION. 2Bl 9. A debt is to be discharged at 11 sereralpayments, m arithmeti- cal series, the first to be $5, and the last ^75 ; what is the whola debt ! the common difference between the several payments ! -^ Ajis, Whole debt, $440 ; common difference, $7 3. What is the sum of the series 1, 3, 5, 7, 9, &c., to 1001 1 Ans. 251001. NoTB. — The number of terms must first be found. 4. A man bought 100 yards of cloth in arithmetical series; he gave 4 cents for tlie ^r5^ yard, and 301 cents for the last yard ; what was the amount of the whole 1 Ans. $152^50. 5. What annuity, in 20 years, at 6 per cent, simple interest, will amount to $1570? Ans. $50. 6. What is the sura of the arithmetical series, 2, 2i|, 3, 3^, 4, 4j^, &c., to the 50th term inclusive ? Ans, 712^. 7. What is the sum of the decreasmg series, 30, 29|, 29|, 29, 28|, &c., down to ? Ans. 1365. 8. A laboring female was able to put $30, at the end of each year, in the savings bank, at 5 per cent., simple interest, from the age of 18 till 77, when she died ; how much had she become worth ? Ans. $4336<50. GEOMETRICAL PROGRESSION. IT 382« 1. A man, having 5 acres of land, doubled the quantity at the end of each year for 4 years; how many acres had he then ? Solution. — Having 5 acres at first, he had 2 times 5, or 10, at the end of the first year, 2 times 10, or 20, at the end of the second year, 2 times 20, or 40, at the end of the third year, and 2 times 40, or 80, at the end of the fourth year. Ans. 80 acres. 2. A lady, having S80, traded at 4 stores, expending one half at the first, half of what she had left at the second, and so on, expending half the money in her possessioi^ at each store till the last : what had she left ? SoLtJTioN. — She leaves the first store, to which she came with -i- 2 = $40, the second with $40 H- 2 =$20, the third with $20 -f.. 2 = $10, the fourth with $10 -f- 2 =$5. Ans. $5. Any series of numbers like 5, 10, 20, 40, 80, increasing by the same multiplier, or like 80, 40, 20, 10, 5, decreasing by the same divisor, is called a geometrical progression. The multiplier or divisor is called the ratio. The. first and last terms are called the extremes. 24* 288 GEOHETBiCAL PROGRESSION. H 223. The first is called an increasing, the second is called a de- creasing geometrical series. Note. — As in arithmetical, so also in geometrical progression^ there an fire things, any three of which being giren, the other tioo may be found : — 1st. The /Krs/ term. 2d. The W term. 3d. The number of terms. 4th. The ratio. 6ih. The sum of all the terms. IT 39S« The first term, number oftermsy arid ratio of an tTicreasing geometrical series being given, to find the last term. 1. A man agreed to pay for 13 valuable houses, worth S5000 each, what the last would amount to, reckoning 7 cents for the first, 4 times 7 cents for the second, and so on, in- creasing the price 4 times on each to the last ; did he gain or lose by the bargain, and how much ? Solution, — "We multiply 7 cents, the sum reckoned for the first house, by 4, to get the sum reckoned for the second house, and this by 4 to get the sum reckoned for the third house, and so on, multiplying 12 times by 4. But to multiply twice by 4 is the same as to multiply once by 16, which is the second power of 4, or 4 times 4. Thus, 7 X ^ s= 28, and 28 X 4 = 112. So, also, 7 X 16= 112. Hence, multiplying 7 by the twelfth power of 4 is the same as multiplying by 4 12 times. And 4«, that is, the twelfth power of 4, is 16777216; and 7X1^777216, (using, if we choose, the larger factor for the multiplicand, Hf 21,) pro- duces 117440512 cents, = $117440542 ; the houses were worth $5000 X 13r=$65000, and $1174405<12~ $65000 s> $110940542, loss, Ans. Hence, when the first term, number of terms, and ratio of an increasing series are given, to find the last term, Multiply the first term by the ratio raised to a power one less than the number of terms. Note 1. — To get a high power of a number, it is convenient to write down a few of th« lower powers, and multiply them together, thus : powers of 4. 1st power. 2d power. 8d power. 4th power. 5th power. 6th power. 7th power. 4 16 64 256 1024 4096 16384, &c. Now the 7th power, multiplied by the 5th power, will produce the 12th power, as also the 6tn by the 4th, and the product by the 2d ; the 5th by the 4tn, and the pniduct bv the 3d ; the 7th by the 4lh, and the product by the 1st, &c Muhii>lyiiig all the powers now written, will produce the 28th ; all but the 5th, will produce the 23d ; all but the 21, will produce the 26lh, &c. Questions* — IT 232* What is a geometricil progression? — an la- creasing scries? — a decreasing series? — the extremes? — the ratiol What nve things are there, of which, if three are given, the others may b% found 12M. OBOMBnUCAL PHOQBB88iO(l 28^ EXAMPIiBS FOR PRACTICB. S. A man plants 4 kernels of corn, which, at harrest, produce 33 kernels : these be plants the second year ; now, supposing the annual increase to continue 8 fold, what would be the produce of the 15th year, allowing 1000 kernels to a pint! Note 2. — The 4 kernels planted is the first term, and the 32 kernels lar- Tested the second term, both within the first year. Ans. 2199023255<552 bushels. 3. Suppose a man had put out one cent at compound interest in 1620, what would have been the amount in 1824, allowing it to double once in 12 years? 2i7» 131072. Ans, $1310*72. NoTS 3. — When the ratio, the number of terms, and the last term of a de- creasing series is given, the first term is evidently found by the same rule. But when the same things are given of a decreasing series, as those of the ascending series required by the rule, that is, the first term, ratio, and number <tf terms, we must divide the first term by the ratio raised to a power which if <nie less than the number of terms. In the same ¥ray we may find the first term of an increasing series, when the ratio, number of terms, and last term are given. 4. If the last term of a decreasing series be 5, the ratio 3, and the number of terms 7, what is the first term t Ans, 3645. 5. If the first term of a decreasing series be 10935, the ratio 3, and the number of terms 8, what is the last term? Ans, 5. 6. If the last term of an increasing series be 196608, the number of terms 17, and the ratio 2, what is the first termi Ans, 3. Note 4. — When the first and last terms, and the ratio are given, to find the number of terms, we majr divide the greater term by the less, the quotient by the ratio, and so on, continually dividing by the ratio till nothing remains • the number of divisions will be equal to the number of terms. 7. The first term is 7, the ratio 10, and the last term 700000000 ; what is the number of terms ? Ans, 9. Compound Interest by Progression. IT S34« 1. To what will $40 amount in 4 years, com* pound interest at 6 per cent. ? Questions* — IT 223* What are given, and what is required, in IT 223 ? How is the last term found, as first described 1 What may be done instead of this? Why? How may a high power be obtained? How the 20tQ power? — the 16th power? —the 27 .h power? — the Uth power? What IS the rule ? Give the substance of no .e 3 ; — of note 4. GEOMETRICAL »OORES8ION 7 224. OPERATION. $40% prm., or Ist term 1*06 240 40 42*40, 2d term. 1*06 25440 42*40 44^9440, 3d term. 1*06 2696640 449440 47*640640, 4th term. 106 285843840 47640640 Solution. — The amount of $40 for one year is once $40-{« j^of $40, Qrl*06 (iU) of $40, and to obtain it, we mul- tiply $40 by 1<06. This pro- duct, multiplied by 1*06, gives the amount for 2 years. Hence, compound interest is a case of an increasing geometrical pro- gression, of which there are given the first term, or princi- pal, the ratio, or the amount of $1 for 1 year, and the number of terms, which is one more than the number of years, there being 2 terms the first year, the principal at the commence- ment, and the amount with one year's interest at the close; and we are to find the last term — the amount for the time — by the rule in the last Hf. The several terms, it will be seen, increase by the common multi- plier, 1^06. Ans, $50*499 -f. Note 1. — The powers of the ratio may usually be foujid in the table, IT 161, since they are the same with the amounts of $l for the number of years which indi- cates the power of the ratio that 50'49907840, 5th term. we wish. It appears also that the only difference in Ending the amount of a sum at compound interest by the table, and by proeression, is the order in which we take the factors. By IT 161, we multiply the amount of $1 for the time by the number of dollars ; by progression, we multiply the number of dollars by the ratio raised to a power denoted by the number of years, or the amount of 91 for the time. EXAMPLES FOR PRACTICE. S. What is the amount of 40 dollars, for 11 years, at 5 per cent., compound interest ? ^rw. $68*413 -|-. 3 What is the amount of $6, for 4 years, at 10 per cent., com- pound interest ? Ans, $8*784^. ,4. In what time will $1000 amonnt to $1191*016, at 6 per cent., compound interest? Note 2. — The case is evidently one of finding the number of terms, (one more than the number of years,) when the ratio and the first and last ieram are given, IT 223, note 4. Afu. 3 years. Questions. — IT 224* How does it appear from the illustration that com* pound interest is a case ofprogression ? What are the things given 7 — to find what ? What is the ratio, Ex. 1 ? Why ? How may tie powers oi the ratio usu- ally be found 7 Why 7 What does it appear are gi\en, and what is required, fiiZ. 47 7226,226. GEOKBTRICAL PROORESBION^ Compound Discount. IT 33S. 1. What is the present worth of $304*899, due 4 years hence without interest, money being worth 6 per cent, compound interest ? Solution. — We find, by the table, ^ 161, that $1, in 4 years, amounts to $ 1*26247, and as many times as this amount of $1 is contained in the given sum, so many dollars it will be worth ; for it is worth a sum, which, put at compound interest 4 years, would amount to it, and divid ing the amount of the number of dollars by the amount of one dollar, — we have the number of dollars, or, Ans. 241^5094-' NoTK. — The case is evidently one of a geometrical progression, in which the ratio, (1'06,) the number of terms, (6,) and the. greater term are given, to find the less, as in IT 223, note 3. TABIiE, Showing the present worth of $1, or £1, from 1 year to 40, allowing compound discount, at 5 and 6 per cent. Years. 6 per cent. 6 per cent. Yean. 5 per cent 6 per cent. 1 *952381 '943396 21 ' ^358942 * '294155 2 *907029 '889996 22 ' ^341850 < '277505 3 *863838 '839619 23 ' '325571 '261797 4 *822702 '792094 24 ' '310068 '246979 5 '783526 '747258 25 ' '295303 '232999 6 '746215 '704961 26 ' '281241 ' '219810 7 '710681 '665057 27 ' '267848 ' '207368 8 '676839 '627412 28 ' '255094 ' 195630 9 '644609 '591898 29 ' 242946 - 184557 10 '613913 '558395 30 ' 231377 - '174110 11 »584679 '526788 3V ' 220359 ^ 164255 12 '556837 '496969 32 ' '209866 ' '154957 13 '530321 '468839 33 ' 199873 - '146186 14 '505068 '442301 34 ' 190355 - '137912 15 '481017 '417265 35 ' 181290 - 130105 16 •458112 '393646 36 ' 172657 * 122741 17 '436297 '371364 37 ' 164436 ^ 115793 18 '415521 '350344 38 ' 156605 - 109239 19 '395734 '330513 39 ' 149148 ' 103056 20 '376889 '311805 40 ' '142046 ' 097222 IT 336. The extremes and the ratio given to find th^ sum of the series. ■■■II III ' ^"~""^~""^ Questions* — IT 225. What is compound discount? How is the pres- ent worth found ? Like what ctuse in a geomeuical progression is it 7 986 GEOMETRICAL PROGRSSSK^f. . f 226 f 1. A man bought 4 yards of cloth, giving 2 cents for the first yard, 6 for the second, 18 for the third, and 54 for the fourth ; what does he pay for all ? SoLVTioM. — We may add together the prices of the several yards thus: 2 + 6 + 18 + 54 :«80. But in a lengthy series, this process would be tedious ; we will therefoiu seek for a shorter method. Writing down the terms of the series, we multiply the first term by the ratio, and place the product over the sec- ond, to which it will be equal, since the second term is the product of the first into the ratio. Multiply, also, the secoud term, placing the product over its equal, the third ; multiply the third, placing the product over the fourth ; multiply the fourth, and place the product at the right of the last product, thus : Second series, 6 18 54 162 First series, 2 6 18 54 162 2 2 ) 160, tioice the first series. 80, sum of the first series. The second series is three times the first series, and subtractmg the first from it, there will remain twice the first series. But the terms bal- ance each other, except the first term of the first series, the sum of which we wish to find, and t^e last term of the second, which is 3 times the last term of the series whose sum we wish. , Subtracting the former from the latter, we have left 160, twice the sum of the first, which dividing by 2, the quotient is 80, sum of the scries required. Hence, RUIiE* Multiply the larger term by the ratio, and subtract the less term from the product, divide the renminder by the ratio less 1 ; the quotient will be the sum of the series. EXAMPLES FOR PRACTICE. 3. If the extremes be 4 and 131072, and the ratio 8, what is the sum of the series ? Ans, 149796. 3. What is the sum of the decreasing series, 3, 1, J , ^, ^V* ^ extended to infinity ? Questions. — IT 22ft. What are ^ven, to find w \at, IT 226 1 How might the sum be found 7 What difficulty in this 7 What is the manner of prooeed« Inff to find a shorter method 7 Give the rule. Explain Uie reason of the ndt What is an infinite series, and what its last temi 1 T227. GfiOBIETBICAL PROGRBSSION. 287 NoTB — Such a teries it calkd an infinite series, the last term of which is so near Dothin; that we regard it ; hence when the extremes are 3 and 0, and the ratio 3, what is the snm of the series? Ans. 4i|. 4. What is the valae of the infinite series, 1 -{- J* -t- iV ~h ^^ &c. ! Ans. 1^. 5. What is the yaloe of the infinite series, -j^^ -|- y^ -|- frftn "h io^oo i &c., or, what is the same, the decimal '11111, &c., continusdly repeated? Ans. j. 6. What is the value of the infinite series, y^ -|- ttt^xr^' ^^'t decreasing by the ratio 100, or, which is the same, uie repeating decimal '020203, &c. t Ans. ^. IT 937* The first term, ratio, and mender of terms given to find the sum of the series. 1. A lady bought 6 yards of silk, agreeing to pay 5 cents for the first yard, 15 for the second, and so on, increasing in a three fold proportion ; what did the whole cost I Solution. — We may find the prices of the several yards, and add them together, or, having foand the last term by ^ 223, we can find the snm bv the last ^. But our object is to find a still more expeditious method. Let us find the several terms and write them down as a first series, and below it write a series which we will call the second, having 1 for thevfirst term, and the same number of terms, thus : First series, 5 15 45 135 405 1215 •th pow«r or ratio. Third series, 3 9 27 81 243 729 Second series, 1 3 9 27 81 243 729 — 1 = 728, which -f- 2 = 364, and 364 X 5.= 1820 cents. Now muUipljring the second series by the ratio, 3, and writing the prod nets as directed in the last Tf, we have a series three limes the second The last term of the third series, it must be carefully noticed, is the 6th power of 3, the ratio, the power denoted by the number of terms. Sub- tracting the second series from the third, which is done by taking 1 from the last term of the third, the other terms balancing, 729 — 1 sss 728, we have twice the second series, and dividing 728 by 2, 728-f-2^364, we have once the second series. Now the first series, the sum of which is required, is 5 times the second, since, as the first term is 5 times greater, each term is 5 times greater than the corresponding tend of the second series ; and multiplying 364, the sum of the second, by 5, we have the required sum, or 1820 cents =■ $18^20, Afu. Hence, the first term, ratio, and number of terms being given, to find the sum of the series, RULE. Raise the ratio to a power whose index is equal to the 288 GEOMETRICAL PEOeBESSK)N. V 226. number of terras, from which subtract 1, and divide the re- mainder by the ratio less 1 ; the quotient is the sum of a series with 1 for the first term; then multiply this quotient by the first term of any required series ; the product will be its amount. EXAMPIiES FOR PRACTICE. 3. A gentleman, whose daughter was married on a new year's day, gave her a dollar, promising to triple it on the first day of each month in the year ; to how much did her portion amount? Applying this rule to the example, 3*^ = 531441, and — ^---r — X 1 = 265720. Ans. $265,720. 3. A man agrees to serve a former 40 years without any other re- ward than 1 kernel of corn for the first year, 10 for the second year, and so on, in tenfold ratio, till the end of the time ; what will be the amount of his wages, allowing lOGO kernels to a pint, and supposing he sells his com at 50 cents per bushel? io«— 1 ^ ,_5 1,111,111,111,111,111,111,111,111, 10—1 ^* — ^ 111,111,111,111,111 kernels. Ans. $8,680,555,555,555,555,555,555,555,555,555,555*555x^1^^. 4. A gendeman, dying, lefl his estate to his 5 sons ; to the young- est $1000, to the second $1500, and ordered that each son should ex- ceed the younger by the ratio of IJ ; what was the amount of the estate? Note. — Before flndmg the power of the ratio IJ, it may be reduced to an improper fraction = ^, or to a decimal, 1*5. \ X 1000 = $13187j; or,-- X 1000 = $13187*50, 2" — 1 1 5 — 1 Answer* Annuities at Compound Interest. IT S828. 1. A man rented a dwelling-house for $100 a year, but did not receive anything till the end of 4 years, when the whole was paid, with compound interest at 6 per cent., on the sums not paid when due"; what did he receive? Questions* -^ IT 22T* What is the first method of finding the sum, when the things are given, named in IT 227 ? — the second ? Describe the process for finding a third method. What terms constitute the first series ? — the second 7 How is the third foimd 7 What do you say of its last term 7 How does it appear to be so 7 How much is the remainder, after subtracting the second from the third series 7 How, then, is the sum of the second series &und7 How the sum of the first, and why 7 Give the rale. Why raise the ratio, tc. 7 • f 229 OEOUETRICAL PROGBESSION. 289 SoLonoH . — As annuities in arrears at simj^e interest lorm an arith- metical series, so the several years' rents with compound interest on those in arrears, are so many terms of a geometrical series. The last year's rent is $100 only, since it is paid when due, at the end of the year ; the third year's rent is on interest 1 year, and is found by multi- ? lying $100 by 1^06, producing $106', and this product multiplied by *06, will give the second year's rent, paid 2 years after it is due, and so on. The first term, $100, the number of terms, 4, and the ratio, VO^ are given to find the sum, as in the last ^, and we may apply the same role, thus : — 1<06<— 1 ^Q^ X 100 » 437*45. Ans. $437*45. NoTB. — The powers of the ratio, see IT 224, may be fofond in the table, Tl61. BXAMPI^ES FOR PRACnCS. 3. What is the amount of an annuity of $50, it being m arrears 20 years, allowing 5 per cent, compound interest? Ans. $1653<29. 3. If the annual rent of a house, which is $150, be in arrears 4 years, what is the amount, allowing 10 per cent, compound interest ! Ans. $696*15. 4. To how much would a salary of $500 per annum amount in 14 years, the money being improved at 6 per cent, compound interest? in 10 years? in 20 years? in 22 years ? in 24 years? Ans, to the last, $25407*75. 5. Two men commence life together ; the one pays cash down, $200 a year to mechanics and merchants ; the second gets precisely tiie same value of articles, but on credit, and proving a negligent pay- master, is charged 20 per cent, more than the other ; what is the diif- ference in 40 years, compound interest being calculated at 6 per cent. ? Ans. $6190*478+. 6. A family removes once a year for 30 years, at an expense and loss of $100 each time ; what is the amount, 6 per cent, compound interest being calculated ? Ans. $7905*818 +. Present Worth of Annuities at Compound Interest. T 339* 1. A man, dying, left to his nephew, 21 years old, the use of a house, which would rent at $300 a year for 10 years, after which it was to come in the possession of his own children ; the young man, wishing ready money to commence business in a small shop, rented the house for 10 l^nestioiia* — IT 238* How does it appear that an annuity is an example of geometrical progression ? Why is the number of terms only equal to the number of years ? What is to be found, and by what rule 7 25 990 GffiOM£TRIGAL FBOGRE29SIOI9. f2sa years, receiving in adratrce such a sum as was equivalent to $300 paid at the end of each year, reckoning compounfd dis- count at 6 per cent. ; what did he receive ? SoLTrrioN. — First, we find what he would receive at the end of 10 3rears, if nothing had been paid before^ by the last If. Now what he should receive at the commencement of the 10 years, is a sum, which, on compound interest at the rate given, would amount to this in 10 years, and we divide it by the amount of $1, found as above, for the pfesent worth. Ans, $2208*024. EXAMPLES FOR PRACTICE* &. What is the present worth of an annual pension of $100, to continue 4 years, allowing 6 per cent, compound interest ^ Atis. $346*503+. 3. What is the present worth of an annual salary of $100, to con- tinue 20 years, allowing 5 per cent. ? Ans. $1246*218 -(-. IT 230* The operations under this rule being somewhat teiious, we subjoin a TABIiE* Showing the present worth of $1 or £1 annuity, at 5 and 6 per cent, compound interest, for any number of years from 1 to 40. feara. Gpereeat, 1 0*95PI38 2 1'85941 3 2*72325 4 3*54595 5 4*32948 6 5'07569 7 5*78637 8 6'46321 9 7*10782 10 7*72173 11 8'30641 12 8*86325 13 939357 14 9*89864 15 10'37966 16 10^83777 17 11*27407 18 11*68958 19 1208532 20 12'46221 6 per cent. Tears. 0*94339 21 1*83339 22 2*67301 23 3*4651 24 4*21236 25 4*91732 26 5*58238 27 6*20979 28 6*80169 29 7*36008 30 7*88687 31 8*38384 32 8*85268 33 9*29498 34 971225 35 10*10589 36 10*47726 37 10*8276 38 1M5811 39 11*46992 40 & per cam. 12»82115 13*163 13*48807 13*79864 14*09394 14*37518 14*64308 14*89813 15*14107 15*37245 15*59281 15*80268 16*00255 16*1929 16*37419 16*54685 16*71128 16*86789 17*01704 17*15908 6 pef oentk 11*76407 12*04158 12*303^ 12*55035 12*78335 13*00316 13*21053 13*40616 13*59072 13*76483 13*92908 14*08398 14*22917 14*36613 14*49824 14*62098 14*73678 14^84601 14*94907 15*04629 I^SSl. GEOMETRICAL MOORESSIQN. S91 NoTB 1. — From the table it appears that, instead oftl a year for 30 years, paid at the end of each, which would be 930, one would receive at the com- mencement, •15'37245, at 5 per cent., or tl3'76483, at 6 per cent, compound discount, and for 950 a year, 60 times as much. Henee, for finding the pres* eat worm at compound discount by the table, Multiply the present worth of SI hy the tininber of dollars. BXAMPId» FOB. PRACTICB. 1. What ready money will purchase an annuity of $150, to con tinue 30 years, at 5 per cent, compound interest? Ans. $S305'8675. 2. What is the present worth of a yearly pension of $40, to cor tinue 10 years, at 6 per cent, compound interest? at 5 per cent. I to continue 15 years I 20 years t 25 years f 34 years! Ans. to last, $647'716. NoTB 2. ~ The practised arithmetieian will have no difficulty in ealcubtinf the present worth of annuities at simple interest, from principles heretofore presented. Annuities at Compound Interest in Reversion. ^ S31 • NoTB. — An annuity b said to be in roversion when it does not commence immediately. 1. In Ex. 1, 1 229, supposing the uncle iiad reserved the use of the house to his sister for 2 years after the young man was 21, and given it to him for 10 years after this time should have expired, how much could he have obtained with which to commence business ? Solution. — If he should wait till he is 23 years old, he could obtain #2208^024, as already found, and he can, at 21, obtain a snm which, at compound interest, would amount, in two years, to $2208*024, or the present worth of this sum paid two years before due, found by ^ 225 to be Ans. $1965a34-. Hence t to find the 'present loorth of an armuity in reversion^ Find the present worth, were it to conimence now, and the present worth of this sum for the time in reversion. Questions. --IT 239. Gi^m the first example. What sum should ha receive now? How is it found? If 230* What appears from the table 7 How is the present worth of tM tiund? Rule. 292 GEOMETRICAL PROGRESSION. IT 232 IQXAMPLBS FOR PRACTICE. 2. What ready money will purchase the reversion of a lease of $60 per annum f to continae 6 years, but not to commence till the end of 3 years, allowing 6 per cent, compound interest to the purchaser ? The present worth, to commence immediately, we find to be $295*039, and ^^^ = 247*72. Ans. $247*72. 3. What is the present worth of $100 annuity, to be continued 4 years, but not to commence till 2 years hence, allowing 6 per cent, compound interest? Ans. $308*392-f- 4. What is the present worth of a lease of $100, to continue 20 years, but no* to commence till the end of 4 years, allowing 5 per cent. ? what, if it be 6 years in reversion 1 8 years ? 10 yearsi 14 years? Ans. to last, $629*426. 5. The revolutionary war closed in 1783 ; one of the soldiers com- menced receiving, in 1817, a pension of $96 a year, which continued till 1640 ; what was the pension worth to him at the close of the war, the rate being 6 per cent, compound interest ? Ans, $162*89 -|-. Perpetual Annuities, IT 333« 1. A fann rents for S60 a year, at 6 per cent ; what is its value ? SoLTTTioN. — This is a perpetual annuity, since the owner is supposed to. receive $60 a year forever. On every dollar which the farm is worth he receives 6 cents, and consequently the farm js worth as many dollars as the number of times 6 cents are contained in $60. $60 -s- $H>6 B$1000, Ans. Hence, to find the worth of a perpetual annuity, Divide the annuity by the rate per cent. ; the quotient will De the perpetual annuity. 2. A city lot is rented 099 years, at $800 a year ; what is it worth, the rate being 7 per cent. ? Note 1. —This is the same as a perpetual annuity. Ana. 811428*67-4^ 3. What is the worth of $100 annuity, to continue forever, allow- ing to the purchaser 4 per cent, ? sdlowing 5 per cent. ? S per cent. 1 10 per cent. 1 16 per cent. ' 20 per cent. • Ans. to last, $500. 4. A farm is left me which will rent for $60 a year, but is . Questions* — IT 231. What do you understand by annuities in rever Hon 7 How is the worth of an annuity in reversion found 7 t233. PERMXTTATION. 293 not to come into my possession till the end of 2 years ; what is it worth to me, the rate being 6 per cent, compound inter* est? % SoLunoK. — The farm will be worth $1000 to me 2 years hence, and it is now worth a sam which, put at compound interest 2 years, wiU amount to $1000. ^r^^^^ z= $889^996. Ans. 5. What is the present worth of a perpetual annuity of $100, to eommence 6 years hence, allowing the purchaser 5 per cent, com- pound interest? wlmt, if 8 years in reversion! 10 years? — 4 years? 16 years? 30 years ? Ans, to last, $462^755. Note 2. — The foregoing^ examples, in compound interest, have been con- fined to yearly payments ; if the payments are Aa^ yearly, we may take ha{f the principal or anmuity^ na\f\h& rate per cent.y and Itpice the number qfyear*^ and work as before, and so for any other part of a year. PERMUTATION. IT 333* Permutation is the method of finding how many different ways the order of any number of things may be va- ried or changed. 1. Four gentlemen agreed to dine together so long as they could sit, every day, in a different order or position; how many days did they dine together ? Solution. — Had there been but two of them, a and J, they could sit only in 2 times 1(1X2 = 2) different positions, thus, a b, and b a. Had there been three, a, b, and c, they could sit in 1X2X3 = 6 differ- ent positions ; for, beginning the order with a, there will be 2 positions, viz., ab Cj and a c b ; next, beginning with b, there will be 2 positions, ba Cj and b c a ; lastly, beginning with c, we have cab, and c b a, that is, in all, 1X2X^ = 6 different positions. In the same manner, if there be/(wr, the different positions will belX2X3X4 = 24. Ans. 24 days. Hence, to find the number of different changes or permzUa' turns, of which any number of dWereni things is capable, — Multiply continucdly together all the terms of the natural Qaestions* — IT 233. What do you understand by a perpetual annuity ? How is its value found 7 Rule. When it does not begin immediately, how is ite worth calculated ? What do you say of other than yearly nayments 1 H 233. What is permutation? Illustra*^ by the first eiample. What if the rule? 25* -^ A»^ . .». r. SM MISCEL];,AI>iBOUS fiXAMPLSa 7234 Beries of numbers, from 1 up to the giveu numbert mi tho last product will be the answer. 3. How many yariations may there be in the position of the aino di^! Ans. 362880. 3. A man bought 05 cows, agreeing to pay for them 1 cent fox every different oider in which they could all be placed ; how much did the cows cost him ? Ans. $1551 181004^330985984000000. t. Christ Church, in Boston, has 8 bells ; how many changes may be rung upon them! Ans. 40320. ^rm MISCELLANEOUS EXAMPLES. ir984. 1. 7+4— 2+3+ 40X5 « how many f iijif.230. Note. — A line drawn oVer seTeral numbers, signifies that the whole are U> be taken as one number. 2. The sum of two numbers is 990, and their difference is 90 ; what are the numbers ? 3. There are 4 sizes of chests, holding respeetively 48, 76, 87 and 90 lbs. ; what is the least number of pounds of tea that will exactly fill some number of chests of either of the 4 sizes ? Ans. 396720 lbs. 4. How many bu^iels of wheat, at 91*50 per bushel^ mpst be given for 15 yards of cloth worth 2s. 3d. sterling per yard ? Atis. 5j^^ bushels. 5. If oats, worth $^0 ner bushel, are sold for $^5 on account, for what ought cloth to be sold on account, worth $3^5 per yard cash ? Ans. $4*374. 5. Bought a book, marked $4*50, at 33 J per cent, discount for cash ; what did I pay ? Ans. $3K)0. 7. Bought 120 gallons of molasses for $42 ; how must I sell it per gallon to gain 15 per cent. ? Ans. $*40|. 8. What sum, at 6 per cent, interest, will amount to $150 in 2 years and 6 months? Ans. $130*434 +. 9. What is the present worth or $1000, pajrable in 4 years and 2 months, discounting at the rate of 6 per cent. I Ans. $800. 10. Bought cloth at $3<50 per yard, and sold it for $4*25 per yard| what did I gain per cent. ? Ans. 21^ per cent. 11. If 20 men can build a bridge in 60 days, how many would be re- quired to build it in 50 days ? Afis. 24 men« 12. How muc)i Silesia, I4 yards wide, will line 12 yards of plaid, | yd. wide? Ans. 5 yards. 13. A cistern, holding 400 gallons, is supplied by a pipe at the rate of 7 gallons in 5 minutes, but 2 gallons leak out in 6 minutes ; in what time will it be filled ? Ans. 6 hours 15 minutes. 14. A ship has a leak which would cause it to sink in 10 hours, but it could be cleared by a pump in 15 hours ; in what time would it sinkf h Ans. 30 hours. 15. How \oxig must I keep $300^ to balsAoe the use of $500, which I lent a friend 4 months ? Ans. 6| months ^ T 234. BOSCRLLANEOUS EXAMPLES. 8M 6. If BOO men hare provisions for 2 months, how many most leave that the remainder may subsist 5 months on the same ? Am. 480. 17. Bought 45 barrds of beef, at $3*50 per barrel, except 16 barrels, for 4 of which I pay no more than for 3 of the others ; what do the wh<^ cost? Ans. $143'50. 18. A hare, rtmning 36 rods a minute, has 57 rods the start of a dog ; how £u must the dog run to ovenake him, running 40 rods per minute? Ans. 570 rods. 19. The hour and minute hands of a watch are together at 12 o'clock ; when are they next together ? Ans. 1 h. 5 m. 27Tnr s. P. M. 20. Three men start together to travel the same way around an island 20 miles in circumference, at the rate of 2^, and 6 miles per hour ; in what time will they be together again ? Ans. 10 hours 21. Two boats, propelled by steam engines 8 miles an hou^start at the same time, the one up, the other down a river, from places 300 miles apart ; at what distance from the place where each started will they meet, if the one is retarded, and the other accelerated 2 mUes an hour by the current ? Ans. 112i| miles from the lower, 187i| from the upper place. 22. The third part of an army were killed, the fourth part taken pris- oners, and 1000 fled ; how many in the army ? Ans. 2400. 23. A fetrmer has his sheep in 5 fields : | in the first, ^ in the second, I in the third, i^ in the fourth, 450 in the fifth ,* how many sheep has he? Ans. 1200. 24. If a pole be J in the mud, f in the water, an4 6 feet out of the water, what is its length? Ans. 90 feet. 25. If iV of a school study grammar, | geography, -^ arithmetic, ^ leam to write, and 9 read, what number in the school ? Ans. 80. 26. A man being asked how many geese he had, replied, if I had } as many as I now have, and 2^ geese more, added to my present num- ber, I should have 100 j how many had he ? Ans. 65. 27. In a fruit orchard, i| the trees bear apples, J pears, -J- plums, 100 peaches and cherries ; how many in all ? ' A7is. 1200. 28. The diflerence between | and f of a number is 6 j required the number. Ans. 80. 29. What number is that, to which, if J and ^ of itself be added, the sum will be 84 ? Ajis. 4%. 30. B's age is IJ times the age of A, C's age 2^ times the age of both^and the sum of their ages is 93 years •, required the age of each. Ans. A 12 years, B 18 years, C 63 years. ^31. If a former had as many more sheep as he now has, i, J, jj, and Jf tt9 many, he would have 435 ; how many has he ? Ans. 120. 32. Required the number, which, being increased by } and f of itself, av-d by 22, will be three times as great as it now is. Ans. 30. 39^ A and B commence trade with equal sums ; A gained a sum equal t6 ^ of his stock, B lost $200. when A's money was twice B's ; what stock had each ? Ans. $500. 34. A man was iired 50 da3r5, receiving S<75 for every da)r he voiced, and forMing 6*25 for every day he was idlej he received $27'50 ; how many days did he work ? Ans. 40. 85. A and B have the same income ; A saves ^ of his ; B, spending 2M MISCELLANEOUS EXAMPLES. IT 234 $30 a year more than A, is $40 in debt at the end of ft years ; what did B spend each year ? -Ans. $205. 36. A man left to A i his property, wanting $20, to B 4, to C the rest, which was $10 less than A*s share ; what did each receive? Ans. A received $80, B $50, C $70. 37. The head of a fish is 4 feet long, the tail as long as the head and i the length of the body, the body as long as the head and tail ; what is the length of the fish ? Ans. 32 feet. 38. A can build a wall in 4 days, B in 3 days ; in what time can both together build it ? Ans. 1^ days. 39. A and B can build a wall in 4 days, B and C in 6 days, A and C in 5 days ; required the time if they work together. Ans. 3-g^ days. 40. A and B can build a wall m 5 days ; A can build it in 7 days ; in how many days can B build it? Ans. 17^ days. 41. A man left his two sons, one 14, the other 18 years old, $1000, so divided, that their shares, being put at 6 per cent, interest, should be equal when each should be 21 years old ; what was the share of each ? ^W5. $546453 + ; $453'846+. 42. What is paid for the rent of a house 5 years, at $60 a year, in. arrears for the whole time at 6 per cent, simple interest? Ans. $336. 43. If 3 dozen pairs of gloves be equal in value to 40 yards of calico, and 100 yards of calico to 90 yards of satinet, wortli $'50 a yard, how many pairs of gloves will $4 buy ? Ans. 8 pairs. 44. A, B, and C divide $100 among themselves, B taking $3 more than A, C $4 more than B ; what is C's share ? Ans. $37. 45. A man would put 30 gallons of mead into an equal number of 1 pint and 2 pint bottles : how many of each ? Ans. 80. 46. A merchant puts 12 cwt. 3 qrs. 12 lbs. of tea into an equal num- ber of 5 lb., 7 lb., and 12 lb. canisters ; how many of each ? Ans. 60. 47. If 18 grs. of silver make a thimble, and 12 pwts make a tea- spoon, how many of each can be made from 15 oz. 6 pwts. ? ^715. 24. 48. If 60 cents be divided among 3 boys so that the first has 3 cents as often as the second has 5 and the third 7, what does each receive ? Ans. 12, 20, and 28 cents. 49. A gentleman paid $18*90 among his laborers, to each boy $«06, to each woman $'08, to each man $46; there were three women for each boy, and 2 men for each woman j how many men were there ? Ans. 90. 50. A man paid $82'50 for a sheep, a cow, and a yoke of oxen j for the cow 8 times, for the oxen 24 times as much as for the sheep ; what did he pay for each? Ans. $2^50, $20, and $60^. 51. Three merchants accompanied ; A furnished f of the capital, B J, and C the rest ; what is C's share of $1250 gain ? Ans: $281*25. 52. A puts in $500, B $350, and C 120 yards of cloth ; they gain $332*50, of which C's share is $120 ; what is C's cloth worth per yard, and what is A's and B's share' of the gain ? Ans. C's cloth $4 per y<l., A's share $125, B's do. $87*50. 53. A, B, and C bought a farm, of which the profits were $580'80 ft year ; A paid towards the purchase $5 as often as B paid $7, and B $4 ^ often as C paid $6 ; what is each one's share of the gain ? Ans. A's share $129*066§, B's $180'693|, C's $271<04. r T234 MISCELLANEOUS EXAMPLES. 297 54. A gentleman divided his fortune among his sons, giving A f9 as often as B $5, and C $3 as often as B $7 ; C received $7442*10A ; whal was the whole estate ? Ans. ^56063'857| . o5. A and B accompany j A put in $1200 Jan. 1st, B put in such a sum, April 1st, that he had half the profits at the end of the year : how much did B put in ? Ans. $1600. 56. Three horses, belonging to 3 men, do work to the amount of $26*45 J A and B's horses are supposed to do | of the work, A and C's ^%, B and C's ^, on which supposition the owners are paid propor- Uonally j what does each receive ? Ans. A $11*50, B $5*75, C $9*20. 57. A ga;r fellow spent ^ of his fortune, after which he gave $7260 for a commission, and continued his profusion till he had only $2178 left, which was | of what he had after purchasing his commission ; what was his fortune ? Ans. $18295*20. 58. A younger brother received £1560, which was -^ of his elder toother's fortune, and 5| times the elder brother's fortune was { of twice as much as the father was worth j what was he worth ? Ans. £19165 14s. 3f d. 59. A gentleman left his son a fortune, -fg of which he spent in 3 months ; | of ^ of the remainder lasted him 9 months longer, when he had only £537 left ; what was the sum bequeathed him by his father? Ans. £2082 18s. 2t^. 60. A general, pladng his army in a square, had 231 men left, which number was not enough by 44 to enable him to add another to each side ; how many men in the army ? Ans. 19000. 61. A militarv officer placed his men in a square j being reinforced . by three times his number, he placed the whole again in a square j again being reinforced by three times his last number, he placed the whole a third time in a square, which had 40 men on each side ; how many men had he at first ? * Ans. 100. 62. Suppose thai a man stands 80 feet from a steeple, that a line to him from the top of the steeple is 100 feet long, and that the spire is three times as high as the steeple ; what is the length of a line reaching from the top of the spire to the man ? Ans. 197 feet nearly. 63. Two ships sail from the same port ; one sails directly east at the rate of 10 miles, the other directlv south at the rate of 7^ miles an hour; how far are they apart at the end of 3 days ? Ans. 900 miles. 64. How many acres in a square field measuring 70*71 rods between the opposite comers ? Ans. 15| acres. 65. Supposing that the river Po is 1320 feet wide and 10 feet deep and runs 4 miles an hour ; in what time will it discharge a cubic mite of water into the sea ? Note. — A linear mile is 5280 feet. Ans. 22 days. 66. If the country which supplies the river Po with water be 380 miles long and 120 "broad, and the whole land upon the surface of the earth be 62,700,000 square miles, and if the quantity of water discharged by the rivers intQ the sea be everywhere proportional to the extent of land by which the rivers are supplied, how many times greater than the Po will the whole amount of the rivers be ? Ans. 1375 times. 67. trpon the same supposition, what quantity of water, altogether} will be discharged by all the rivers into the sea in a year of 365 days 1 Ans. 22S124 cnl»c milM. 298 MEASUREMENT. H 235 68. If the proportion of sea to land be as lOi to 5, and the average depth of the sea oe 1^ miles, in how long time, if the sea were empty would it be filled? Ans. 8657 years 275 days. 69. If a cubic foot of water weigh 1000 oz., and mercury be 13^ times heavier than water, and the hight of the mercury in the barometer (which weighs the same as a column of air on the same base and ex- tending to the top of the atmosphere) be 30 inches, what will the air weigh on a square foot ? — on a square mile ? What will the whole atmosphere weigh ? Ans.f in order, 2109*375 lbs., 5880»M)00000 lbs., 11430122220000000000 lbs. 70. A traveler who had set a perfectly accurate watch by the sun in Boston, 71° 4' W. Ion., being in Detroit, 82° 58' W. Ion., 3 days after, was surprised to find it wrong, when compared with the sun j was it too fast or too slow? how much, and why ? 71. A building fell in Portland, Me., 70° 20^ W. Ion., at 9 o^clock, A. M., and in 3 minutes the intelligence of the event reached St. Louis, Mo., 90° 15^ W. Ion., by magnetic telegraph j when was it knbwn at St. Louis? Arts. At 43 m. 20 sec. past 7 o'clock, A. M. 72. At the battle of Bunker Hill the roar of cannon was distinctly heard at Hanover, N. H, and business was suspended for a time ; in what time did the sound pass, the distance being supposed 120 miles ? Note. — Sound moves 1142 feet in a second. Ana. 9 m. 14 sec. +• 73. Seeing the flash of a rifle in the evening, it was 8 seconds before I heard the report j what was the distance ? Ans. 1 mi. 3856 ft. 74. A man in view on a hill opposite is chopping, at the rate of a blow in 2 seconds ; I saw him strike 4 blows before I heanl the first ; what is his distance from me ? Ans. 1 mi. 1572 ft. 75. A laborer dug a cellar, the length of which was 2 times the width, and the width 3 times the depth ; he removed 144 cubic yards of earth ; what was the length ? .^5. 36 feet. 76. A owes B $750, due in 8 months; but receiving $300 ready money, he extends the time of paying the remainder, so that B shaU lose nothing; when must it be paid? Ans. In 1 yr. 1 mo. 10 days. 77. The sum of two numbers is 266§, and the product of the greater multiplied by 3, equals the product of the less multifdied by 5 ; what are the numbers? ilns. 100, and 166§. 78. A park 10 rods square is surrocmded by a walk which occupies -^t^ of the whole park ; what is its width ? Ans. 8 ft. 3 in. 79. A, B and Cf x^mmence trade with $3053*25, and gain $610*65 ; A*s stock -|- B's, is to B*s -|- C's, as 5 to 7 ; and C's stock — B*s, is to C*s 4- B's, as 1 to 7 ; what is each one's part of the gain ? Ans. A's gain $135*70, B's $203*55, C's $271*40. MEASUREMENT OF SURFACES. IT 33tS« To find the area of a parallelogram^ multiply the length by the shortest distance hetween the sides. T23& MEASDBEMENT M9 Note 1. — A parallelcgmm has its opposite sides equal, but its adja<.ent sides unequal, like the figure A B C D, or E F C D. The former is callecT a rectangle, see IT 48. The second is called a rhomlioid. and is equal in size to the first. 1. What are the superficial content'* '^f an oblique angled piece of ground, measuring 80 rods in length auu 90 rods in a perpendicular fine between its sides f Ans: 1600 sq. rods. Note 2. — To find the contents o. ? rhombus, which, like the annexed figure, has its sides equal, but its angles not right angles : multiply the length of one side by the shortest distance to the side opposite. To JiTid the area of a trapezoid, multiply half the sum of the parallel sides by the shortest distance between them, NoTK 3. — A trapezoid is a figure, like the one in the an* nexed diagram, bounded by four straight lines, only two of which are parallel. 2. What is the area of a piece of ground in the form of a trape- zoid, one of whose parallel sides is 8 rods, the other 12 rods, and the perpendicular distance between them 16 rods ? A+ia X 16 == 160 sq. rods, Ans, 3. How many square feet in a board 16 feet long, 1^8 feet wide at one end, and 1*3 at the other 1 Ans. 24*8 feet. To find the area of a triangle^ multiply the bctse by half the altitude. NoTB 4. — The figure ABC is a trian- §le, of which the side A B is the base, D C le altitude. The triande b evidently half the parallelogram A B C F, the area of which equals A B X 1^ C. A D B 4. The base of a triangle is 30 rods, and the perpendicular 10 rods ; what is the area? Ans. 150 rods. 5. If the contents are 600 rods, and the base 75 rods, what is the altitude ? Ans, 16 rods. 6. Required the base, the area being 400, the altitude 40 rods Ans. 20 rodfc 7. How many square feet in a board 18 feet long, li feet wide at one end, and running to a point at the other 1 Ans, 13^1 feet. To find the circumference of a circle lohen the diameter is knottm, multiply the diameter by 3^, or accurately by 3*14159. To find the area, multiply the circumference by one fourth of the diimeter • or multiply the square «>f the diameter by •7854 300 MEASUREMENT. If 236 Note 5. — The principles on which the above ni'©» ave fixHicled) as well aa those for the measurement of many figures, will be understood by a geometrical demonstration, and the pupil must take them without a demonstration till he may study that interestmg science. 8. What is th(B circumference of a circular pond, the diameter of which is 147 feet? What is the areal Ans. to the last, 1697 Ixxr feet. 9. If the circumference be 22 feet, what is the diameter? Ans, 7 feet. 10. If the diameter of the earth is 7911 miles, what is the circum- ference? Ans, 24853 miles. 11. How many square inches of leather will cover a ball 3;^ inches m diameter? Note 6. — The area of a ball is 4 times the area of a circle having the same cCiameter. Ana. 38i square inches. 12. How many square miles on the earth's surface? Ans, 196,612,083. Measurement of Solids. V 339« NoTB 1 . — The general principle for finding the contents of solid bodies is to multiply the length by the breadth, and the product by the thickness, but the rule applies directly only to the cube or right prism, being subject to modifications as applied to solid figures of other forms. See *l 61. 1. How many solid inches in a globe 7 inches in diameter t Note 2. — The solid contents of a globe are found by multiplying the area of its surface by ^ part of its diameter, or the cube of its diameter by '523^. Ans, 179$ solid inches. 2. What number of cubic miles in the earth ? V Ans, 259,233,031,435i. 3. What are the solid contents of a log 20 feet long, of uniform size, the diameter of each end being 2 feet? •Note 3. — A fi^re like the above is called a cylinder. To find the solid contents, we find the area of one end by a foregoing rule, and multiply the area thus found by the length. Ans, 62*83 4-cu. ft. 4. A bushel measure is 18*5 inches in diameter, and 8 inches deep; iiow many cubic inches, does it contain ? ^725. 2150*4 -^. N'»TB 4. — Solids having bases bounded by straight lines, and decreasing unilormly till the^f come to a point, are called pyramids. Solids which thus decrease, with eireular bases, are called cones. Pyramids and cxxaeB are jost one third as large as cylinders, of which the area of each end is equal to the area of the bases of these solids. Hence, if we multiply the area of the base by the hight, and divide the product by 3, the quotient will be the solid con- tents. 5. What are thQ solid contents of a pyramid, the baao of which i0 4 feet square, and the perpendicular hight 9 feet? Am, 48 solid fael. f S37, 238. HEASUItEMENT. 901 0. What ftie the solid contents of a cone^ ^e faight of whidi il 27 ieet, and ^ diameter of the base is 7 feet ? Ans, 3461 solid feet. 7. What are the solid contents of a stick of timber 18 feet long, one end of which is 9 inches square and the other end 4 inches square, uniformly diminishing throughout its wh(^ length ? Note 5. — Such a figure is called the fhistum of a pyramid, and the solid contents are found by adding to the areas of the ends the square root of their product, and multipiyiDg the sum by one third of the hight. The pu^il must notice that the diameters are ezpreased in inches, while the length is m feet. An$, 5 solid feet, 936 solid indies. 8. What are the solid contents of a round log of wood, 36 feet long« 1*6 feet in diameter at one end, and diminishing gradually to a diaoieter of *9 of a foot at the o^er ? NoTB 6. — Such a fi^re is called the frustum of a cone, and the solid con- tents are found by adding to the squares of the two diameters the square root of the product of those squares, multiplying the sum by *7854, and the result* ing product by one third of the length. Am. 45*333 -f* solid feet. Oanging, of Measuring Casks. V 3ST» 1. How many gallons of wine will a cask contun, tilie head diameter of which is 25 inches, and the bung diameter 31 inches, and the length 36 inches? How many beer gallons ? Note. — Add to the head diameter 2 thirds, or if the staves curve but slightly, 6 tenths of the difference between the head and bune diameters, the sum will be the average diameter. iTie cask will then be reduced to a cylin- der, the contents of which may be found by a foregoing rule, in solid inches. The solid Inches may be divided by 231 j (IT 114,) to find the number of wine gallons which the cask will contain, and by 282, (IT 115,) to find the number r gallons. ^^ i03«93 -J- wine gallons, 84*32 4" ^^^ gallons. 2. How many wine gallons in a cask, the bung diameter of whicb is 36 inches, the head chameter 27 inches, and the length 45 inches? Ans, 166*617. Mechanical Powers. T 388* 1. A lever is 10 feet long, and the fulcrum, or prop, on which it turns is 2 feet from one end ; how many pounds weight at the short end will be balanced by 42 pounds at the other end 1 Note 1. — In turning rwind the prop, the long end will evidently pass over a space of 8 inches, while the short end passes over a space ot 2 inches. Now. it IS a fundamental principle in mechanics, that the weight and power will exactly balance each other y when they are inversely as the spaces they pass over. Hence, in this example. 3 pounds, 8 feet from the prop, will balance 8 pounds 2 feet from the prop ; tnerefore, if we divide the distance of the powsb Jrom the prop by the distance of the weight yrom the prop^ the quotient will always express the ratio of the weight to the poweb ; ^ =s 4, that is, the meS^ht will be 4 time» as much as the power. 42 X 4 — 168. Am. 168 lbs. 26 -*n 302 MBASUBEMENT. IT 9. Supposing the lever as aboTS, what power would it require to raise 1000 pounds 1 Ans. JLi^=:250 pounds. 3. If the weight to be raised be 5 times as much as the power to be applied, and the distance of the weight from the prop be 4 feet, how far from the prop must the power b« applied ? Ans. 20 feet. 4. If the greater distance be 40 feet, and the less ^ of a foot, and the power 175 pounds, what is the weight t Ans. 14000 pounds. 5. Two men carry a kettle, weighing 200 pounds ; the kettle is suspended on a pole, the bale being 2 feet 6 inches from the hands of one, and 3 feet 4 inches from the hands of the other ; how many pounds does each bear? . ( 114^ pounds. ' ( 8&f pounds. 6. Hiere is a windlass, the wheel of which is 60 inches in diame- ter, aiKl the axis, around which the rope coils, is 6 inches in diameter; now many pounds on the axle will be balanced by 240 pounds at the wheel? Note 2. — The 8pace» passed 0T«r are as the diameten^ or the fxrornifer* eneea; therefore, -^ =s 10, ratio. Ans. 2400 pounds. 7. If the diameter t>f the wheel be 60 inches, what must be the diameter of the axle, that the ratio of the weight to the power may be 10 to 1 ! Ans. 6 inches. Note 3. — This calculationis on the supposition, that there is no/rtc/wm, for which it is usual to add i to the power wnich is to work the machine. 8. There is a screw, the threads of which are 1 inch asunder ; if it is turned by a lever 5 feet, =b60 inches, long, what is the ratio of the weight to the power! Note 4. — The power applied at the end of the lever will describe the cir- cumference of a circle 60X^ = 120 inches in diameter, whUe the weight is raised 1 inch : therefore, the rcUio will be found by dividing- the eircum/erenee of a circle^ whose diameter is twice the length of the lever f by the distance be- tween the threads of the screw. 3774- 120 X 3| = 377| circumference, and — I = 377f , ratio, Ans, 9. There is a screw, whose threads are \ of an inch asunder ; if it be turned by a lever 10 feet long, what weight will be balanced by 120 pounds power % Ans. 362057f pounds. 10. There is a machine, in which the power moves over 10 feet, while the weight is raised 1 inch ; what is the power of that machine, that is, what is the ratio of the weight to the power ? Ans. 120. 11. A man put 20 apples into a wine gallon measure, which was afterwards filled by pouring in 1 quart of water ; required the con- tents of the apples in cubic inches. Ans. 173^ inches. 12. A rough stone was put into a vessel, whose capacity was 14 wine quarts, which was afterwards filled with 2^ quarts of water ; what was the cubic content of the stone? Ans. 664 1 inches. Note 6. — For a more full consideration of the foregoing subjects, the pupil ts referred to the forthooming treatise on Mensoration in conneetion with ths "flsriea.*' FORMS OF NOTES» ETC. 303 FORMS OF NOTES, RECEIPTS, AND ORDERS. Notes. No. I. Koene, Sept. 17, 1846. For Talue reoeived, I promise to pay to Olivbe Bountiful, or order, sixty- Jiree dollars, fifty-foor cents, on demand, with interest after three mouths. Attest, Timothy Tsstimont. William Trusty. No. n. Ludlow, Sept. 17. 1346. For Talue received, I promise to pay to O. R., or bearer, dollars seats, three months after date. Pbtbb Pencil. No. III. By ttpo persona, Yates, Sept. 17, 1846. For Talne received, we, jointly and severalljr, promise to pay to C. D., or <wder, dollars cents, on demand, with interest. Attest, pBTfiB Saxs. Aldem Faithful. James Fairface. Receipts. Boston, Sept. 19, 1846. Recelted from Mr. Dubancb Adley ten dollars in full of all account^ Orvand Constance. Newark, Sept. 19, 1846. Received of Mr. Obvand Constance five dollars in full of all accounts. Durance Adley. Receipt far Money received on a Note. Rochester, Sept. 19, 1946. Received of Mr. Simpson Bastey (by the hand of Titus Trusty) sixteen dollars twenty-five cents, which is endorsed on his note of June 3, 1846. Peter Cheerful. A Receipt for Money received on Account'. Hancock, Sept. 19, 1846. Received of Mr. Orland Landikb fifty dollars on account. Eldro Slackley. Receipt Jor Money received Jbr another Person. Salem. August 10, 1846. Received from P. C. one hundred dollars tor account of J. B. Eli Truman. Receipt Jor Interest due on a Note. Amherst, July 6, 18461 Received of I. S. thirty doUiirs, in full of one year's interest of 9500, due to OQ the day of last, on note from the said I. S. Solomon GhtAV. 304 POKMS OP BILLS. Receipt for Money paid before it becomes due. HUlsbonrngh, ISttf 8, ld99. Received of T. Z. ninety dollars, advanced in full for one year's tent oF my fiurm, leased to the said T. Z., ending the first day of April next, 1147. HONMTUS JaMSI. NoTB. — There is a distinction between receipts ^ven in full of all aecourU8f and others in full of all demanda. The former cut offaccounU only; the lat- ter cut off not only aeooiints, but all obligations and right of action. Orders, Utica, Sept. 9, 1846. Mr. Stbphkn Btraoiss. FV>r value reeeiwd, pay to A. B., or order, ten dollars, and place the same to my account. Samuel Skiunsb. Pittsburgh, Sept. 9, 1846. Mr. Jambs Robottom. Please to deliver to Mr. L. D. such goods as he mav call for, not exceeding the sum of twenty-five dollars, and place the same to tne account of your humble servant, Nicholas Rbubbks. FORMS OF BILLS. " Before you bmldf sit down and count the cost,** ^Iheon Thrifty built a house for Thomas Payw^ll, according to a plan agreed upon between them, for the sum of tl500. The cellar of the houre is 24 by 2d feet, and is dug 4 feet deep below the top of the ground. The cellar walls are 7 feet high. There is a wing at one end of the main building, 20 by 24 feet, which is underpinned with a wall 3 feet high. As one side of the wing joins tne main building, for the underpinning of it but 3 walls are required, one 24, and two 20 feet long on the outside. The walls of the cellar, and the un- derpinning of the wing are li feet thick. To the cellar there is a door 4 by 7 feet, j^nd 2 windows, each 2 by 2^ feet. Simeon Thrifty, wishing to know how much it cost him to build the house, kept an accurate account of all the materials used, the labor employed, and tne cost of each. The folk>wing are his bills : — BUI of Timber, 6 sticks for posts to upright part, each 14 ft. long, and 4 by 10 iit 6 " " wing, "11 « 4 " 10 " 2 " sills to upright part, .....," 28 " 7 " 8 ** 6 " " and sleepers to upright and wing, ........ 24 " 7 " 8 " 4 " plates and side girts to upright. "28 " 6 " 7 " 7 " " and beams to u]m^t ana wing, "24 « 6 " 7 " 2 " eave gutters, "30 " 6 " W " 3 " " "20 " 6 " id " 62 rafters, "16 " 8 " 4 " 96 studs, "12 " 2 " 4 " 176 partition planks, • * 10 " 2 " 4 " 96 scantlings, " 12 " 4 " 4 " 10 " for braces, "12 ** 4 " 4 " 76 joists, "14 " 9"T" ae" •'10 •■ a"f* FORMS OF BILLS. 305 « BUI <(f Lumber. 800 ft. best qoality pine, for best doors, and other nice joiner work. 10000 " common ** " door and window casings, stairs, base boaidi oonmion doors, &e., dc 8850 ** white wood siding. 3160 " bass " aoormg. 160C roof boards. / * 6000 " lath. / 26 bunches shingles, 500 in each bunch. ' A. Bm<(fMaienaiafor Windewt. > Sash and ^ass for 14 windows of 24 panes each, 7 jy9 inches. V "• « 4 " " 20 « 7 «' » " ^, (- M i( 2 " " 6 ** 7 " ft " / " " 1 window, " 16 " 7 " » « « u 1 " " 12 " 7 '* 9 " 90 lbs. putty. Hardware BiU, 4 casks nails, 100 lbs. each. 22 pairs 3 inch door hinges, with screws. 2 « 4 « <( « 20 door handles. " 2 outside door Knobs and locks. 3 cupboard fastenings. 63 ft. tin eave conductors, including 4 elbows. 3 stove-pipe crocks. 3 " thimbles. 3 fl. tin pipe for sink-spout. 20 window springs. * 4 papers 1 mch brads. BUI of MaieridUfor Oiimneys and Plattervnff. 1600 bricks. 27 loads sand. 200 bush. lime. 10 bush, liair. BiU ofPrieee of Materials, Stone for cellar walls and underpinning, . . . . t *26 per perch. All the timber except the eave gutters reduced to board measure, that is, 1 inch thick, ... 10^ " M. Eaveguuers, 15* " " Pine lumber, best quality, 20* « ** ♦* " common, .10* " " Siding, flooring, and roof boards, 10* ^* ** Lath, . ■; 5* « " Shingles 1'60 ** bunch. Wmdowsash, *03 ** pane. ** "g^ass, 2*50 '* bozoflUpanMk Putty, *07 ** lb. Nails, *05i« « 8 inch door butts, with screws, 42i ** pdr. Doorhandles, *12ieach. Outside door knob^and locks, 1^50 ** Cupboard fastenings, '12} " Eave conducters, '12} per ft. Extra for elbows, '06i each. Stove-pipe crocks, ••• '37i *< « '^ '^ thimbles, ^2* « 26* 306 VOBJeS OF BILLS. Sink spout, ^ ..... t '44. Window ^ring^ *06i each. Bnuis, 40 per paper. Bricks, 10* « M. Lime, <12J « bush. Sand, 1<00 " k>ad. Hair, <26 « bush. BUI of Prices of Labor. Digging 'cellar, t '18 per cu, yd. 6 stone masons, 6 days saoh, |9 2'00 2 carpenters, 12 " ** 1'50 8 joiners, 40 " " 1*76 Painting and glazing. lOOfOO. Furring ready for lathing, . 12'00. Lathing, , 15*00. 2 plasterers, 7 days each, . . 19 2*50 per day. 8 brick-layers, 1 day " « 2*76 ** Team and hired man, 4 monUis of 26 days each, ....** 2'00 ** Simeon Thrifty commenced the house on the 1st day of^ay, and completed it on the 3d day of S^t. s allowing him $1*60* per week for his board, how much did he get for his own labor ? Arts. 9281 *64<^. (I day. fC 76-^ perches stone, . . 600 ft. eave gutters, . . 8287 " other timber, . . 800 " best pine lumber, 17610 " other lumber, . . 6000 " lath, 26 bunches shingles, . Window sash, 4 boxes window glass, . 30 lbs. putty, 400 " nails, Butt hinges, Door hanidles. Outside door Jmobs & lockS) Cupboard fastenings, . . Tin eare conductors, . . . E^tra on elbows, .... Stove-pipe crocks, .... ** thimbMSi . • • Sink spout, ....... Window spring!, * . • • • 18'88-]2^ 9*00 82*87 16'00 176*10 30*00 39'00 13<68 10*00 2'10 22'00 3<06 2*50 8*00 *37i 7*874 *26 1*12J 'Z7i *44 1*26 Brads, 1600 bricks. 200 bush, lime, . . . 27 loads sand, . . . 10 bush, hair, . . . Dig^ng cellar, . . . Laying stone work, . Carpenters' work, . . Joiners' ** . , Painting and glazing, Furring, ....... Lathing, ...... Plastenn^, Bnck-laying, . . . . Team and hired man, 18 weeks' boaid^ . . « I *40 9*60 25*00 27*00 2*50 4*48 60*00 36*00 210*00 100*00 12*00 15*00 35*00 7*60 208*00 27*00 Amount, . . • . tl218*35^l| Errors excepted, J. n. F. SCHOOL BOOK DEP OS TORY. PHILLIPS &. SAMPSON, PU8LI8HERS, BOOKSELLERS, AND STATIONERS, 110 WASHinGTon Stbbst, {Up StairSt) BOSTON, Keep constantly on hand one of the best assortments of SCHOOL BOOKS, Stationery, &c. to be found in the country. COUNTRY BOOKSELLERS, TOWNS, TRADERS, SCHOOL COMMITTEES, TEACHERS, and others purchasing* School Books, are respectfully requested to call and examine the Stock of SCHOOL BOOKS and Stationery in this establishment. Particular attention is paid to furnishing all the various kinds of School Books now in use in the Col- leges, Academies, and Schools in the country, on the most favorable terms. Also, a very extensive assortment of Standard, Theo- logical, Claasicad and Miscellaneous Books, which will oe sold at prices as reasonable as can be had in the COUNTRY. STATIONERY, BLANK ACCOUNT BOOKS, Paper, Quills, tik. Slates, &c. of every variety. Orders solicited and promptly attended to. u l< u It tt tt u tt Tbomas Sbbkwin, a M., Principal of the English High School, Boaton. Bawtom Field, " " Fraaldin " " Samitbl S. Gruns, PHDCiptl of the Phillips School, BoetoiL Joshua Batbs, Jr., *' ** Brimmer JosLiH A. Stbarns, " " Mather Uaao F. Shbpard, " " Otis' GaoRas B. Htdb. " " Dwight J. D. Philbrick, Principal of the Mathematical Depar't of Mayhew School, Boaton. WuxiAM A. Sbbpard, " " " ^' " " " D. P. Paob, ** " New York Sute Normal School, Albany. P. H. SwBBTSBR, " " Harvard School, Charlestown. ELBRnmB Smith, '' ** Classical and English High School, Worcester. C. C. DsANS, '* " English High School, Newburyport. Hon. Francis Bwioht, late of Albany, N. Y. Jamxs D. Batchbldbr, Principal of uiammar School, Marblebead. Charlbs Awards, " " " " Academy at Concord, Mass. Union Seminary, Fairhareo. Brown School, Salem. Hacker « Pickering " Phillips " Epes ** Fish " Salstonstall " Grammar SclMxd, Lynn. « tt ft tt tt tt tt K tt tt tt tt It tt tt U tt Charlxs W. Goodmow, Alonzo Tripp, J. B. Fairfibld, D. P. Gaixoup, Albert Laokbt, Joseph Williams, Charles Northend, Oliver Carlton, Edwin Joceltn, Jacob Batchelobe, Jr., William S. Wiluamb, Amasa Davenport, Elwell Woodburt, William T. Adams, John Capen, Isaac Swan, George Newoomb, Teaclier of Qainey Orammaiir ScKod. E. Wtman, Principal of the English and Classical High School, St. Louis, Ma Charles A. Lord, A. M., late Professor in Marion College, Ohiow Z. Grover, Principal of Prospect street School,') ^Tmdtun of the Dorchester Gnmmar Schools. tt tt «. tt tt tt tt tt tt tt tt tt Provi d encSi B. L L. B. Nichols, " " Arnold J. D. GiDDiNOs, " " Fountain CFarnum, " " Elara AmosPbrrt, " " Summer S. S. Ashley, " " Meeting C. T. Keith, " " Benefit Ebenbzbr Harvbt, ) Benjamin Evans, > Principals of Grammar Schools, New Bedford. Cteus Baetlbtt, ) # COMMON SCHOOL AI^EBRA, by Thomas Sherwin, A. M., Principal of the English High School, Boston ; author of Elementary Treatise on Algebra, oc This Algebra is now used as the text book in the Public Schools of Boston, Box- bury, Salem, ftc ; also, in the Phillips' Academy, AndoTer; Bradford Academy, Bradford ; Windsor Academy, Windsor, Yt. ; Fitchburg Academy, and numeroos other places ; and has been recommended by the Superintendents of Public Schools for the whole state of Rhode Island. A large number of testimonials from our best practical teachers are in the hands of the publishers, from wliich they would take ibe following : Phillips School, March 13, 1846 Mr. Sheewxn, —Sir : I hare examined vour " Common School Algebra," and beliere it better adapted to the wants of the banner tiian any otlier I hare seen. Hie plan of introducing preliminary exercises, to acquaint the pupil with the use of letters, is a happy one. By tlie aid of tlieae introductory lessons, the transition from arithme- tic to dgebra is made so easy ttiat a child of ordinary capacity would meet with little difficulty in determining the use of x, v, and x, even without a teacher. I hope to the book extensively used, in that ckss of schools for w hich i t was prepared. SAMUEL & GBElENE. Grammar Master of the PkOUpe School RoxBUET, Fetaraary 6, 1846. Mr. Shbrwin, — Dear Sir : I have been highly gratified by the examination of yoar "Common School Algebra." The principles of the science are unfolded and explidned with great perspicuity and simplicity. I think your reasonings and illustrations are -L-l 49 peculiariy happy and appropriate; and, on the whole, I consider it luperior to any work of the kind that I have erer Men. With much esteem, Your obclient serranti ___ LEVI REED. RozBuRT, February 5, 1846. I PT7LLT concur in the opinion of Mr. Reed in regard to Mr. Sherwin's " Common School Algebra." I hare introduced it into my school, and shall esteem it a privilege to recommend its use whenever an opportunity presents. JEREMIAH PLYMPTON. MiTHXir SoBO<»., Boeton, February 12, 1846. Thomas Shirwut, Esq.,— Dear Sir : I have examined the " Common School Alge- bra," which you sent me, and think it better adapted to the wants of the schoolroom than any other book, uptm the same subject, with which I am acquainted. Yours very truly, WILLIAM D. S\7AN. Boston, December 8, 1845. Thomas Shsrwoi, Eso.,— My Dear Sir : I hare examined your " Common School Algebra," which yon kindly sent me, with great satisfaction. It seema to be a very natural and lucid development of the principles of computation by algebraic symbols, and makes the wtiole subject appear much more practical than it does in most works of this description. Indeed, I think you hare succeeded remarkably in removing what you justly call the ** great difficulty in the studv of algebra," by helping the pupil to "a clear comprehension of the eariieet steps." Ttie work throughout bears the marlcs of an experience in teaching, as well as theoretical familiarity with tlie science elucidated. Accept of mj very sincere thanki^ and assurance of the truest esteem and affection. H. WINSLOW. Chhtral Plaoh, Boston, February 4, 1846. Thomas Sbkbwin, Eso.,— My Dear Sir : I have examined your " Common ScIkx>1 Algebra " with much pleasure. It seems to me so admirably adapted for a first book in ttie science, that I snail adopt it for my next class. I like it, because it appears in every part to be the work of a practical teacher, who has observed Uie wants of his pupils, and skilfully contrived to meet them. Very truly yours, SOLOMON ADAMS. Beimmbr School, January 3l8t, 1846. Thomas Shbrwin, Esq., — Dear 9r: I have carefully examined your "Common School Alfiebra," and I talce pleasure in expressing to you my opinion of its merits. I beliere ft to be the iiest elementary treatise upon Algebra that has ever been pub* lisiied for tlie use of our scImwIs. Strictly scientific, it is yet entirely adapted to the ' capacity of that class of scholars for whom it was designed : and it combines very just proportions of simplicity and strengto; Tlw selection or subjects is sufficiently extensive to render it lueftil to those who are Jesirous of pursuinof a higher course of mathenuttical studv, without being too severe for the comprehension of young pupls ; and the careful and iudicious arrangement of tlie progressive exercises cannot foil to reduce it to the level of less than medium ability. As a practical teacher, I heartily thank you for your efforts to facilitate the acquisition of this important sci- ence. Yours with respect, WILLIAM A. SHEPARD. Mr. SHSswm's " Common School Algebra " I regard as a work of superior merit, and consider it decidedly prefers^le to any other elementary text-book, upon the same subject, with which I am acquainted. Its chief peculiarity, it not fts highest excel- lence, is to be found in the " preliminary exercises," which open a new and delight- ful avenue to this branch of matliematical sciance. They are admirably calculated to communicate to the learner the first principles of Algebra in the most simple and intelligible manner. Having made use of ttem in my class last year, I can speak of their excellence with confidence. In a wsrd, the book is just the thingwe need in our schools. Yours, &c., J. D. PHILBRICK. Matbbw School. GRAMMATICAL CHART OF SENTENCES, by Samuel & Greene, A. M., Principal of Phillips School, Boston. The above /hart contains a classification and illustration of all the component parts of everv sentence in the English language, with definitions, directions af.d models, to aid thw teacher in using it. It is desig^d to assist in teaching the grammar of the language on a more philosophical plan ; it exhibits the structure of sentences, by commencing with the simplest forms, and advancing gradually througli every variety of construcuon, and will ha found a very valuable help in the tedious process of teachins grammar. L 4»V In Preaa, aM wUl Bhortfy be pudtithed, A GRAMIMAR. ON AN ENTTRELY NEW PLAN. By & S. Greene, A. M., Principal of the Phillips School, Boston, which is intended to acconmanj the above | Chart. Hie pubKlshers feel warranted in saying, (from the opinions ofour best teach- ers who have seen thQ proof-sheets,) that the above work will need only to be seen to be universally adopted ; — that it will open to the pupil new light on this hitherto dry study. FOLSOM'S LIVY. TIti Livii Paiavina Historiarum Liber Primus et Selecta queedam Capita. Curavit Notulisque instruzit, Carolus Folsom, Academia fiar- vardiansB olim Bibliothecarius. 15th Stereotype Edition. IK>LLEN'S PRACTICAL GRAMMAR OF THE GERMAN LANGUAGR 12th edition. POLLEN'S GERMAN READER, fur Beginners, llth edition. FoUen's German Grammar and Reader have been very highly reccunmeiided. The second and third editions of the Grammar wsre both greatly improved, by corrections and additional rules and illustrations, by the author; since which, the work has passed through several editions, and has been introduced into Harvard University. The German Reader also, has been repeatedly printed ; and its popularity con- stantly increasing. A NATURAL HISrORY of the most remarkable Quadrupeds^Birdfl, Fishes, Serpents. Reptiles, and Insects. By Mrs. Mary Trimmer. With 200 Engravinga Abridged and improved ; particularly designed for youth of the United States, and suited to the use of Schools. 19th edition. MEADOWS' FRENCai AND ENGLISH PRONOUNCING DICnONARY. A new French and English Pronouncing Dictionary, on the basis of Nu^ent's, with manjr new words in general use. In two parts. First, Frencli and English ; second, English and French, exhibiting the pronunciation of the French in pure English sounds: the parts or speech, gender of French nouns, regular and irregular conjuga- tions of verbs, accent of English w6rds, list of the usual Christian and proper names, and names of countries and nations, to which are prefixed, principles of French pro- nunciation, and an abridged grammar. By F. C. Meadows, M. A., of the University of Paris. New edition, revised and improved by Charles L. Parmentier, A. M., Pro- fessor of the French Laoiguage and Literature. I8mo., sheep. LE BRUN'S TELEMAQUE. Les Aventures de T^l^maque, fils d'Ulysse. Tax M. P^n^km. Edited hy Mr. Charles Le Brun. 1 voL 12mo., sheep. THE COMMON SCHOOL SONG-BOOK. By Asa Fitz, author of the American School Song Book, Primary Song Book, ice. THE SABBATH SCHOOL MINSTREL. By Asa Fitz, author of the Common School Song-Book, &c, being a collection of pleasant Hymns set to Music, and adapted to the use of Sabbath Schools. MODERN GEOGRAPHY AND ATLAS. By J. E. Worcester; new edition. ANCIENT CLASSICAL AND SCRIPTURAL GEOGRAPHY AND ATLAS. By J. E. Worcester. ELEMENTS OF PLANE GEOMETRY. By N. H. Tillinghast, Principal of tUe Bridgewater Normal SchooL VALUABLE BOOKS FOR LIBRARIES. LIVES OF THE HEROES OF THE AMERICAN REVOLUTION, comprising the Life of Washington and his Genenrals and Officers who were most distinguished in the War of the Independence of the United States of America. Also, the Constitution of the United Stales and Amendments, the Declaration of ladepeodenoe, with Wash- ington's Inaugural, First Annual, and Farewell Addresses, ^bellisbed wfth Por- traits. I vol., 12mo. LIFE AND CAMPAIGNS OF NAPOLEON BONAPARTE, giving an account of all hi3 engagements from the siege of Toulon to the battle of Waterloo. Also, em- bracing accounts of the daring exploits of his Marshals, together with his public and private life, from the commencement of his career to his final imprisonment and death on St. Helena, translated from the most authentic soiirces. Ulustrated with fine engravings. I vol., I2mo. BANCROFT'S LIFE OF GEORGE WASHINGTON, being the most impartial and authentic biography of this illustrious patriot that has been heretofore published ; also, embracing a History of the Wars of tho American Revolution. I vol., I2ma, Illustrated. &3- t»ini4JPS & SAMPSON keep constantly on hand one of the largest aasort- ments of SCHOOL, CLASSICAL and MISCELLANEOUS BOOKS, STATIONERY, *c. , to bo found in New England, which will be sold at prices as low as at any other b ookstor e m the country. 48 •/ . ' / I ■ / / . ' ' / / > // ■'■ / \ .' \ y ' s ./" V * s / . / :-/•', / V '/ / P I I ' i ^ r- I *• / - ■J ''. ( \ ■■ < , «• / . x ■ / / / / ^ <r / "\ / ' / X / - / 5- *•— ■— '\-'- r% ^ I f , /, / /^-/-/;-A -.- '^' ^' \ ■ / ■U / /■ / / .^loi / '> ^ 7 ^w^n- >> //f- / • r ■ / '^- 7 r~ ■\f ^3k 0r J- h ■* -. J "■ / -1 i - ;> f ^7 . ' / : Z y a' .V r//-4A/^-w-c-. ^^ / /i' \^. i X£^^ z>_J.^ ( . ' •^/// '-'■^ W J/ .■ -'1 r ■1 ^ i / \ \ \ / ^ / '--■ - € 7 f/ / \ • I aiUMf 4 * 1 YB 35797 ' ' ' ' ' ' , , . , • ' . t ' * * 4 \'^ c K" / n xjOl / _v.^ 984153 THE UNIVERSITY OF CALIFORNIA LIBRARY '< t. < J. 4 (-' •;,-2^ ■ C' <^ ^^' / //^ „ 2^'- ' . V/^'-^